The number of elements in the set $\left\{x \in\left[0,180^{\circ}\right]: \tan \left(x+100^{\circ}\right)=\tan \left(x+50^{\circ}\right) \tan x \tan \left(x-50^{\circ}\right)\right\}$ is $\_\_\_\_$ .
Explanation:
$ \begin{aligned} &\begin{aligned} & \tan \left(x+100^{\circ}\right)=\tan \left(x+50^{\circ}\right) \tan x \tan \left(x-50^{\circ}\right) ; x \in\left[0,180^{\circ}\right] \\ & \frac{\tan \left(x+100^{\circ}\right)}{\tan x}=\tan \left(x+50^{\circ}\right) \tan \left(x-50^{\circ}\right) \\ & \frac{\sin \left(x+100^{\circ}\right) \cdot \cos x}{\cos \left(x+100^{\circ}\right) \cdot \sin x}=\frac{\sin \left(x+50^{\circ}\right) \sin \left(x-50^{\circ}\right)}{\cos \left(x+50^{\circ}\right) \cos \left(x-50^{\circ}\right)} \end{aligned}\\ &\text { Apply Componendo and Dividendo rule }\\ &\begin{aligned} & \frac{\sin \left(x+100^{\circ}\right) \cos x+\cos \left(x+100^{\circ}\right) \sin x}{\sin \left(x+100^{\circ}\right) \cos x-\cos \left(x+100^{\circ}\right) \sin x}=\frac{\sin \left(x+50^{\circ}\right) \sin \left(x-50^{\circ}\right)+\cos \left(x-50^{\circ}\right) \cos \left(x+50^{\circ}\right)}{\sin \left(x+50^{\circ}\right) \sin \left(x-50^{\circ}\right)-\cos \left(x+50^{\circ}\right) \cos \left(x-50^{\circ}\right)} \\ & \text { use } \sin A \cos B \pm \cos A \sin B=\sin (A \pm B) \text { and } \cos A \cos B \pm \sin A \sin B=\cos (A \mp B) \\ & \frac{\sin \left(x+100^{\circ}+x\right)}{\sin \left(x+100^{\circ}-x\right)}=\frac{\cos \left[\left(x+50^{\circ}\right)-\left(x-50^{\circ}\right)\right]}{-\cos \left[\left(x+50^{\circ}\right)+\left(x-50^{\circ}\right)\right]} \\ & \frac{\sin \left(2 x+100^{\circ}\right)}{\sin 100^{\circ}}=\frac{\cos 100^{\circ}}{-\cos 2 x} \end{aligned} \end{aligned} $
$ \begin{aligned} & \frac{\sin \left(90^{\circ}+\left(2 x+10^{\circ}\right)\right)}{\sin \left(90^{\circ}+10^{\circ}\right)}=\frac{\cos \left(90^{\circ}+10^{\circ}\right)}{-\cos 2 x} \\ & \frac{\cos \left(2 x+10^{\circ}\right)}{\cos 10^{\circ}}=\frac{-\sin 10^{\circ}}{-\cos 2 x} \\ & \cos \left(2 x+10^{\circ}\right) \cos 2 x=\sin 10^{\circ} \cos 10^{\circ} \\ & 2 \cos \left(2 x+10^{\circ}\right) \cos 2 x=2 \sin 10^{\circ} \cos 10^{\circ} \\ & 2 \cos \left(2 x+10^{\circ}\right) \cos 2 x=\sin 20^{\circ} \\ & \cos \left(4 x+10^{\circ}\right)+\cos 10^{\circ}=\sin 20^{\circ} \\ & \cos \left(4 x+10^{\circ}\right)=\sin 20^{\circ}-\cos 10^{\circ} \\ & \cos \left(4 x+10^{\circ}\right)=\sin 20^{\circ}-\sin 80^{\circ} \quad\left\{\cos 10^{\circ}=\sin \left(90^{\circ}-10^{\circ}\right)\right\} \\ & =2 \cos 50^{\circ} \sin \left(-30^{\circ}\right) \\ & =-2 \cos 50^{\circ} \times \frac{1}{2}=-\cos 50^{\circ}=\cos \left(180^{\circ}-50^{\circ}\right) \\ & \cos \left(4 x+10^{\circ}\right)=\cos 130^{\circ} \end{aligned} $
$ \begin{aligned} & 4 x+10^{\circ}=2 n \pi \pm 130^{\circ}, n \in I \\ & 4 x+10^{\circ}=360^{\circ} n \pm 130^{\circ}, n \in I \\ & 4 x+10^{\circ}=360^{\circ} n+130^{\circ} \text { or } 4 x+10^{\circ}=360^{\circ} n-130^{\circ} \\ & 4 x=360^{\circ} n+120^{\circ} \text { or } 4 x=360^{\circ} n-140^{\circ} \end{aligned} $
$ \begin{aligned} &\begin{aligned} & x=\frac{360^{\circ} n+120^{\circ}}{4} \\ & n=0, x=30^{\circ} \\ & n=1, x=\frac{480^{\circ}}{4}=120^{\circ} \\ & n=2, x=\frac{360^{\circ} \times 2+120^{\circ}}{4} \\ & x=210^{\circ} \end{aligned}\\ &\begin{aligned} & x=\frac{360^{\circ} n-140^{\circ}}{4} \\ & n=1, x=\frac{360^{\circ}-140^{\circ}}{4}=\frac{220^{\circ}}{4}=55^{\circ} \\ & n=2, x=\frac{360^{\circ} \times 2-140^{\circ}}{4}=\frac{580^{\circ}}{4}=145^{\circ} \\ & n=3, x=\frac{360^{\circ} \times 3-140^{\circ}}{4}=\frac{940^{\circ}}{4}=235^{\circ} \end{aligned} \end{aligned} $
for $x \in\left[0,180^{\circ}\right]$ solutions are
$ x=30^{\circ}, 55^{\circ}, 120^{\circ}, 145^{\circ} $
so, total number of solutions are $=4$ Ans
The number of solutions of $\sin ^2 x+\left(2+2 x-x^2\right) \sin x-3(x-1)^2=0$, where $-\pi \leq x \leq \pi$, is ________.
Explanation:
$\begin{aligned} & \sin ^2 x+\left(3-(x-1)^2\right) \sin x-3(x-1)^2=0 \\ & \sin ^2 x+3 \sin x-(x-1)^2 \sin x-3(x-1)^2=0 \\ & \left.\sin x(\sin x+3)-(x-1)^2\right)[\sin x+3]=0 \end{aligned}$
There are two intersections between this graph.
So, Number of solution will be 2 .
Let $S=\left\{\sin ^2 2 \theta:\left(\sin ^4 \theta+\cos ^4 \theta\right) x^2+(\sin 2 \theta) x+\left(\sin ^6 \theta+\cos ^6 \theta\right)=0\right.$ has real roots $\}$. If $\alpha$ and $\beta$ be the smallest and largest elements of the set $S$, respectively, then $3\left((\alpha-2)^2+(\beta-1)^2\right)$ equals __________.
Explanation:
For real roots
$\begin{aligned} & D \geq 0 \\ & \sin ^2 2 \theta \geq 4\left(\sin ^4 \theta+\cos ^4 \theta\right)\left(\sin ^6 \theta+\cos ^6 \theta\right) \end{aligned}$
Put $\sin ^2 2 \theta=t$
$\begin{aligned} & \Rightarrow t \geq 4\left(1-\frac{t}{2}\right)\left(1-\frac{3 t}{4}\right) \\ & 2 t \geq(2-t)(4-3 t) \\ & 3 t^2-12 t+8 \leq 0 \\ & t^2-4 t+\frac{8}{3} \leq 0 \\ & (t-2)^2+\frac{8}{3}-4 \leq 0 \\ & (t-2)^2 \leq \frac{4}{3} \\ & -\frac{2}{\sqrt{3}} \leq t-2 \leq \frac{2}{\sqrt{3}} \\ & 2-\frac{2}{\sqrt{3}} \leq t \leq 2+\frac{2}{\sqrt{3}} \\ & \because t \in[0,1] \\ & \Rightarrow 2-\frac{2}{\sqrt{3}} \leq t \leq 1 \\ & \alpha=2-\frac{2}{\sqrt{3}}, \beta=1 \\ & \Rightarrow 3\left[(\alpha-2)^2+(\beta-1)^2\right]=4 \end{aligned}$
If m and n respectively are the numbers of positive and negative values of $\theta$ in the interval $[-\pi,\pi]$ that satisfy the equation $\cos 2\theta \cos {\theta \over 2} = \cos 3\theta \cos {{9\theta } \over 2}$, then mn is equal to ____________.
Explanation:
$ \begin{aligned} & \cos \frac{5 \theta}{2}+\cos \frac{3 \theta}{2}=\cos \frac{15 \theta}{2}+\cos \frac{3 \theta}{2} \\\\ & \cos \frac{5 \theta}{2}=\cos \frac{15 \theta}{2} \\\\ & \frac{15 \theta}{2}=2 n \pi \pm \frac{5 \theta}{2} \\\\ & \frac{15 \theta}{2} \pm \frac{5 \theta}{2}=2 n \pi \\\\ & 10 \theta=2 n \pi \quad \text { or } 5 \theta=2 n \pi \\\\ & \theta=\frac{n \pi}{5} \text { or } \theta=\frac{2 n \pi}{5} \\\\ & \Rightarrow \theta=\frac{n \pi}{5} \end{aligned} $
$ \begin{aligned} & \theta=\pm \pi, \pm \frac{4 \pi}{5}, \pm \frac{3 \pi}{5}, \pm \frac{2 \pi}{5}, \pm \frac{\pi}{5} \\\\ & m=5, \quad n=5 \\\\ & m n=25 \end{aligned} $
Let $\mathrm{S = \{ \theta \in [0,2\pi ):\tan (\pi \cos \theta ) + \tan (\pi \sin \theta ) = 0\}}$. Then $\sum\limits_{\theta \in S} {{{\sin }^2}\left( {\theta + {\pi \over 4}} \right)} $ is equal to __________.
Explanation:
possible values are $\mathrm{n}=0,1$ and $-1$ because
$ -\sqrt{2} \leq \sin \theta+\cos \theta \leq \sqrt{2} $
Now it gives $\theta \in\left\{0, \frac{\pi}{2}, \frac{3 \pi}{4}, \frac{7 \pi}{4}, \frac{3 \pi}{2}, \pi\right\}$
So $\sum \limits_{\theta \in \mathrm{S}} \sin ^2\left(\theta+\frac{\pi}{4}\right)=2(0)+4\left(\frac{1}{2}\right)=2$
Let $S=\left\{\theta \in(0,2 \pi): 7 \cos ^{2} \theta-3 \sin ^{2} \theta-2 \cos ^{2} 2 \theta=2\right\}$. Then, the sum of roots of all the equations $x^{2}-2\left(\tan ^{2} \theta+\cot ^{2} \theta\right) x+6 \sin ^{2} \theta=0, \theta \in S$, is __________.
Explanation:
$7{\cos ^2}\theta - 3{\sin ^2}\theta - 2{\cos ^2}2\theta = 2$
$ \Rightarrow 4\left( {{{1 + \cos 2\theta } \over 2}} \right) + 3\cos 2\theta - 2{\cos ^2}2\theta = 2$
$ \Rightarrow 2 + 5{\cos ^2}\theta - 2{\cos ^2}2\theta = 2$
$ \Rightarrow \cos 2\theta = 0$ or ${5 \over 2}$(rejected)
$ \Rightarrow \cos 2\theta = 0 = {{1 - {{\tan }^2}\theta } \over {1 + {{\tan }^2}\theta }} \Rightarrow {\tan ^2}\theta = 1$
$\therefore$ Sum of roots $ = 2({\tan ^2}\theta + {\cot ^2}\theta ) = 2 \times 2 = 4$
But as $\tan \theta = \, \pm \,1$ for ${\pi \over 4},{{3\pi } \over 4},\,{{5\pi } \over 4},\,{{7\pi } \over 4}$ in the interval $(0,\,2\pi )$
$\therefore$ Four equations will be formed
Hence sum of roots of all the equations $ = 4 \times 4 = 16.$
Let $S=\left[-\pi, \frac{\pi}{2}\right)-\left\{-\frac{\pi}{2},-\frac{\pi}{4},-\frac{3 \pi}{4}, \frac{\pi}{4}\right\}$. Then the number of elements in the set $\mid A=\{\theta \in S: \tan \theta(1+\sqrt{5} \tan (2 \theta))=\sqrt{5}-\tan (2 \theta)\}$ is __________.
Explanation:
Let $\tan \alpha = \sqrt 5 $
$\therefore$ $\tan \theta = {{\tan \alpha - \tan 2\theta } \over {1 + \tan \alpha \tan 2\theta }}$
$\therefore$ $\tan \theta = \tan (\alpha - 2\theta )$
$\alpha - 2\theta = n\pi + \theta $
$ \Rightarrow 3\theta = \alpha - n\pi $
$ \Rightarrow \theta = {\alpha \over 3} - {{n\pi } \over 3}\,\,\,\,\,\,\,\,\,;\,n \in Z$
If $\theta \in [ - \pi ,\,\pi /2]$ then $n = 0,1,2,3,4$ are acceptable
$\therefore$ 5 solutions.
If the sum of solutions of the system of equations $2 \sin ^{2} \theta-\cos 2 \theta=0$ and $2 \cos ^{2} \theta+3 \sin \theta=0$ in the interval $[0,2 \pi]$ is $k \pi$, then $k$ is equal to __________.
Explanation:
Equation (1) $~~~2{\sin ^2}\theta = 1 - 2{\sin ^2}\theta $
$ \Rightarrow {\sin ^2}\theta = {1 \over 4}$
$ \Rightarrow \sin \theta = \, \pm \,{1 \over 2}$
$ \Rightarrow \theta = {\pi \over 6},{{5\pi } \over 6},{{7\pi } \over 6},{{11\pi } \over 6}$
Equation (2) $~~~2{\cos ^2}\theta + 3\sin \theta = 0$
$ \Rightarrow 2{\sin ^2}\theta - 3\sin \theta - 2 = 0$
$ \Rightarrow 2{\sin ^2}\theta - 4\sin \theta + \sin \theta - 2 = 0$
$ \Rightarrow (\sin \theta - 2)(2\sin \theta + 1) = 0$
$ \Rightarrow \sin \theta = {{ - 1} \over 2}$
$ \Rightarrow \theta = {{7\pi } \over 6},{{11\pi } \over 6}$
$\therefore$ Common solutions $ = {{7\pi } \over 6};\,{{11\pi } \over 6}$
Sum of solutions $ = {{7\pi + 11\pi } \over 6} = {{18\pi } \over 6} = 3\pi $
$\therefore$ $k = 3$
Let ${S_1} = \{ x \in [0,12\pi ]:{\sin ^5}x + {\cos ^5}x = 1\} $
and ${S_2} = \{ x \in [0,8\pi ]:{\sin ^7}x + {\cos ^7}x = 1\} $
Then $n({S_1}) - n({S_2})$ is equal to ______________.
Explanation:
Given,
${S_1} = \left\{ {x \in \left[ {0,12\pi } \right],\,{{\sin }^5}x + {{\cos }^5}x = 1} \right\}$
${S_2} = \left\{ {x \in \left[ {0,8\pi } \right],\,{{\sin }^7}x + {{\cos }^7}x = 1} \right\}$
(1) ${\sin ^5}x + {\cos ^5}x = 1$
This satisfies when $\sin x = 1$ and $\cos x = 0$
$\therefore$ $x = {\pi \over 2},{{5\pi } \over 2},{{9\pi } \over 2},{{13\pi } \over 2},{{17\pi } \over 2},{{21\pi } \over 2}$
It also satisfies when $\sin x = 0$ and $\cos x = 1$
$\therefore$ $x = 0,\,2\pi ,4\pi ,6\pi ,8\pi ,10\pi ,12\pi $
$\therefore$ Accepted values of x in $\left[ {0,12\pi } \right]$ is = 13
$\therefore$ $n({S_1}) = 13$
(2) ${\sin ^7}x + {\cos ^7}x = 1$
This satisfies when $\sin x = 1$ and $\cos x = 0$
For $x \in \left[ {0,\,8\pi } \right]$, possible values
$x = {\pi \over 2},{{5\pi } \over 2},{{9\pi } \over 2},{{13\pi } \over 2}$
It also satisfies when $\sin x = 0$ and $\cos x = 1$
$x \in \left[ {0,8\pi } \right]$, possible values
$x = 0,2\pi ,4\pi ,6\pi ,8\pi $
$\therefore$ Total accepted values of x in $\left[ {0,8\pi } \right]$ is = 9
$\therefore$ $n({S_2}) = 9$
$\therefore$ $n({S_1}) - n({S_2}) = 13 - 9 = 4$
The number of solutions of the equation $\sin x = {\cos ^2}x$ in the interval (0, 10) is _________.
Explanation:
$\sin x = {\cos ^2}x$, $x \in (0,10)$
$ \Rightarrow \sin x = 1 - {\sin ^2}x$
$ \Rightarrow {\sin ^2}x + \sin x - 1 = 0$
$\therefore$ $\sin x = {{ - 1 \pm \,\sqrt {1 + 4} } \over 2}$
$ \Rightarrow \sin x = {{ - 1 \pm \,\sqrt 5 } \over 2}$
We know $\sin \in ( - 1,1)$
$\therefore$ ${{ - 1 - \sqrt 5 } \over 2}$ can't be a value of sin x
$\therefore$ $\sin x = {{\sqrt 5 - 1} \over 2}$
$3\pi = 3 \times 3.14 = 9.42 < 10$
${{7\pi } \over 2} = {7 \over 2} \times 3.14 = 10.99 > 10$
$\therefore$ 10 will be in between 3$\pi$ and ${{7\pi } \over 2}$.
There are 4 intersection at A, B, C and D between sin x graph and $y = {{\sqrt 5 - 1} \over 2}$ graph.
$\therefore$ possible solution = 4
The number of elements in the set $S = \{ \theta \in [ - 4\pi ,4\pi ]:3{\cos ^2}2\theta + 6\cos 2\theta - 10{\cos ^2}\theta + 5 = 0\} $ is __________.
Explanation:
$\Rightarrow 3 \cos ^{2} 2 \theta+\cos 2 \theta=0$
$\Rightarrow \cos 2 \theta=0$ or $\cos 2 \theta=\frac{-1}{3}$
As $\theta \in[0, \pi], \cos 2 \theta=\frac{-1}{3} \Rightarrow 2$ times
$\Rightarrow \theta \in[-4 \pi, 4 \pi], \cos 2 \theta=\frac{-1}{3} \Rightarrow 16$ times
Similarly, $\cos 2 \theta=0 \Rightarrow 16$ times
$\therefore $ Total 32 solutions
The number of solutions of the equation
$2\theta - {\cos ^2}\theta + \sqrt 2 = 0$ in R is equal to ___________.
Explanation:
Given,
$2\theta - {\cos ^2}\theta + \sqrt 2 = 0$
$ \Rightarrow 2\theta + \sqrt 2 = {\cos ^2}\theta $
$ \Rightarrow 2\theta + \sqrt 2 = {{1 + \cos 2\theta } \over 2}$
$ \Rightarrow 4\theta + 2\sqrt 2 = 1 + \cos 2\theta = y$ (Assume)
$\therefore$ $y = 4\theta + 2\sqrt 2 $ and
$y = 1 + \cos 2\theta $
For $y = 1 + \cos 2\theta $
when $\theta = 0$, $y = 1 + 1 = 2$
when $\theta = {\pi \over 4}$, $y = 1 + \cos {\pi \over 2} = 1$
$\theta = {\pi \over 2}$, $y = 1 + \cos \pi = 1 - 1 = 0$
For $y = 4\theta + 2\sqrt 2 $
when $\theta = 0$, $y = 2\sqrt 2 $
when $\theta = {\pi \over 2}$, $y = 2\pi + 2\sqrt 2 $
$ = 2(\pi + \sqrt 2 )$
$ = 2(3.14 + 1.41)$
$ = 2(4.55)$
$ = 9.1$
when $\theta = - {\pi \over 2}$, $y = - 2\pi + 2\sqrt 2 $
$ = 2( - \pi + \sqrt 2 )$
$ = 2( - 3.14 + 1.41)$
$ = - 3.46$
$\therefore$ Two graph cut's at only one point so one solution possible.
The number of values of x in the interval $\left( {{\pi \over 4},{{7\pi } \over 4}} \right)$ for which
$14\cos e{c^2}x - 2{\sin ^2}x = 21 - 4{\cos ^2}x$ holds, is ____________.
Explanation:
${{14} \over {{{\sin }^2}x}} - 2{\sin ^2}x = 21 - 4(1 - {\sin ^2}x)$
Let ${\sin ^2}x = t$
$ \Rightarrow 14 - 2{t^2} = 21t - 4t + 4{t^2}$
$ \Rightarrow 6{t^2} + 17t - 14 = 0$
$ \Rightarrow 6{t^2} + 21t - 4t - 14 = 0$
$ \Rightarrow 3t(2t + 7) - 2(2t + 7) = 0$
$ \Rightarrow {\sin ^2}x = {2 \over 3}$ or $ - {7 \over 3}$ (rejected)
$ \Rightarrow \sin x = \, \pm \,\sqrt {{2 \over 3}} $
$\therefore$ $\sin x = \, \pm \,\sqrt {{2 \over 3}} $ has 4 solutions in $\left( {{\pi \over 4},\,{{7\pi } \over 4}} \right)$
Explanation:
${\sin ^4}\theta + {\cos ^4}\theta - \sin \theta \cos \theta = 0$
$ \Rightarrow 1 - {\sin ^2}\theta {\cos ^2}\theta - \sin \theta \cos \theta = 0$
$ \Rightarrow 2 - {(\sin 2\theta )^2} - \sin 2\theta = 0$
$ \Rightarrow {(\sin 2\theta )^2} + (\sin 2\theta ) - 2 = 0$
$ \Rightarrow (\sin 2\theta + 2)(\sin 2\theta - 1) = 0$
$ \Rightarrow \sin 2\theta = 1$ or $\sin 2\theta = - 2$ (Not Possible)
$ \Rightarrow 2\theta = {\pi \over 2},{{5\pi } \over 2},{{9\pi } \over 2},{{13\pi } \over 2}$
$ \Rightarrow \theta = {\pi \over 4},{{5\pi } \over 4},{{9\pi } \over 4},{{13\pi } \over 4}$
$ \Rightarrow S = {\pi \over 4} + {{5\pi } \over 4} + {{9\pi } \over 4} + {{13\pi } \over 4} = 7\pi $
$ \Rightarrow {{8S} \over \pi } = {{8 \times 7\pi } \over \pi } = 56.00$
$|\cot x| = \cot x + {1 \over {\sin x}}$ in the interval [ 0, 2$\pi$ ] is
Explanation:
$\cot x = \cot x + {1 \over {\sin x}} \Rightarrow $ not possible
Case II : When cot x < 0, $x \in \left[ {{\pi \over 2},\pi } \right] \cup \left[ {{{3\pi } \over 2},2\pi } \right]$
$ - \cot x = \cot x + {1 \over {\sin x}}$
$ \Rightarrow {{ - 2\cos x} \over {\sin x}} = {1 \over {\sin x}}$
$ \Rightarrow \cos x = {{ - 1} \over 2}$
$ \Rightarrow x = {{2\pi } \over 3},{{4\pi } \over 3}$(Rejected)
One solution.
Explanation:
$ \Rightarrow $ $\sqrt 3 {\cos ^2}x - \sqrt 3 \cos x + \cos x - 1 = 0$
$ \Rightarrow \sqrt 3 \cos x(\cos x - 1) + (\cos x - 1) = 0$
$ \Rightarrow (\cos x - 1)(\sqrt 3 \cos x + 1) = 0$
$\cos x = 1$
$ \Rightarrow x = 0$ $ [as x \in \left[ {0,{\pi \over 2}} \right]$]
and $\cos x = - {1 \over {\sqrt 3 }}$ (not possible in $x \in \left[ {0,{\pi \over 2}} \right]$]
$ \therefore $ Number of solution = 1
Number of solutions of $\sqrt{3} \cos 2 \theta+8 \cos \theta+3 \sqrt{3}=0, \theta \in[-3 \pi, 2 \pi]$ is :
5
4
3
0
The number of solutions of the equation
$ \cos 2\theta \cos \frac{\theta}{2} + \cos \frac{5\theta}{2} = 2\cos^3 \frac{5\theta}{2} $ in $ \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] $ is :
5
7
6
9
The number of solutions of the equation
$(4-\sqrt{3}) \sin x-2 \sqrt{3} \cos ^2 x=-\frac{4}{1+\sqrt{3}}, x \in\left[-2 \pi, \frac{5 \pi}{2}\right]$ is
If $\theta \in[-2 \pi, 2 \pi]$, then the number of solutions of $2 \sqrt{2} \cos ^2 \theta+(2-\sqrt{6}) \cos \theta-\sqrt{3}=0$, is equal to:
The sum of all values of $\theta \in[0,2 \pi]$ satisfying $2 \sin ^2 \theta=\cos 2 \theta$ and $2 \cos ^2 \theta=3 \sin \theta$ is
Let $|\cos \theta \cos (60-\theta) \cos (60+\theta)| \leq \frac{1}{8}, \theta \epsilon[0,2 \pi]$. Then, the sum of all $\theta \in[0,2 \pi]$, where $\cos 3 \theta$ attains its maximum value, is :
If $2 \sin ^3 x+\sin 2 x \cos x+4 \sin x-4=0$ has exactly 3 solutions in the interval $\left[0, \frac{\mathrm{n} \pi}{2}\right], \mathrm{n} \in \mathrm{N}$, then the roots of the equation $x^2+\mathrm{n} x+(\mathrm{n}-3)=0$ belong to :
The sum of the solutions $x \in \mathbb{R}$ of the equation $\frac{3 \cos 2 x+\cos ^3 2 x}{\cos ^6 x-\sin ^6 x}=x^3-x^2+6$ is
If $\alpha,-\frac{\pi}{2}<\alpha<\frac{\pi}{2}$ is the solution of $4 \cos \theta+5 \sin \theta=1$, then the value of $\tan \alpha$ is
If $2 \tan ^2 \theta-5 \sec \theta=1$ has exactly 7 solutions in the interval $\left[0, \frac{n \pi}{2}\right]$, for the least value of $n \in \mathbf{N}$, then $\sum_\limits{k=1}^n \frac{k}{2^k}$ is equal to:
The number of elements in the set
$S=\left\{\theta \in[0,2 \pi]: 3 \cos ^{4} \theta-5 \cos ^{2} \theta-2 \sin ^{6} \theta+2=0\right\}$ is :
Let $S=\left\{x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right): 9^{1-\tan ^{2} x}+9^{\tan ^{2} x}=10\right\}$ and $\beta=\sum_\limits{x \in S} \tan ^{2}\left(\frac{x}{3}\right)$, then $\frac{1}{6}(\beta-14)^{2}$ is equal to :
The number of elements in the set $S=\left\{x \in \mathbb{R}: 2 \cos \left(\frac{x^{2}+x}{6}\right)=4^{x}+4^{-x}\right\}$ is :
Let $S=\left\{\theta \in\left(0, \frac{\pi}{2}\right): \sum\limits_{m=1}^{9} \sec \left(\theta+(m-1) \frac{\pi}{6}\right) \sec \left(\theta+\frac{m \pi}{6}\right)=-\frac{8}{\sqrt{3}}\right\}$. Then
Let $S=\left\{\theta \in[0,2 \pi]: 8^{2 \sin ^{2} \theta}+8^{2 \cos ^{2} \theta}=16\right\} .$ Then $n(s) + \sum\limits_{\theta \in S}^{} {\left( {\sec \left( {{\pi \over 4} + 2\theta } \right)\cos ec\left( {{\pi \over 4} + 2\theta } \right)} \right)} $ is equal to:
The number of solutions of $|\cos x|=\sin x$, such that $-4 \pi \leq x \leq 4 \pi$ is :
Let for some real numbers $\alpha$ and $\beta$, $a = \alpha - i\beta $. If the system of equations $4ix + (1 + i)y = 0$ and $8\left( {\cos {{2\pi } \over 3} + i\sin {{2\pi } \over 3}} \right)x + \overline a y = 0$ has more than one solution, then ${\alpha \over \beta }$ is equal to
The number of solutions of the equation
$\cos \left( {x + {\pi \over 3}} \right)\cos \left( {{\pi \over 3} - x} \right) = {1 \over 4}{\cos ^2}2x$, $x \in [ - 3\pi ,3\pi ]$ is :
Let $S = \left\{ {\theta \in [ - \pi ,\pi ] - \left\{ { \pm \,\,{\pi \over 2}} \right\}:\sin \theta \tan \theta + \tan \theta = \sin 2\theta } \right\}$.
If $T = \sum\limits_{\theta \, \in \,S}^{} {\cos 2\theta } $, then T + n(S) is equal to :
$2\cos x\left( {4\sin \left( {{\pi \over 4} + x} \right)\sin \left( {{\pi \over 4} - x} \right) - 1} \right) = 1,x \in [0,\pi ]$ and S is the sum of all these solutions, then the ordered pair (n, S) is :
${{\cos x} \over {1 + \sin x}} = \left| {\tan 2x} \right|$, $x \in \left( { - {\pi \over 2},{\pi \over 2}} \right) - \left\{ {{\pi \over 4}, - {\pi \over 4}} \right\}$ is :
1 + sin4 x = cos23x, $x \in \left[ { - {{5\pi } \over 2},{{5\pi } \over 2}} \right]$ is :