Trigonometric Equations

1 - 2sin²x + $\alpha $ sin x = 2$\alpha $ - 7

$ \Rightarrow $ 2sin²x - $\alpha $ sin x + (2$\alpha $ - 8) = 0

$\sin x = {{\alpha \pm \sqrt {{\alpha ^2} - 8(2\alpha - 8)} } \over 4}$

$ \Rightarrow {{\alpha \pm \sqrt {{\alpha ^2} - 16\alpha + 64} } \over 4} = {{\alpha \pm (\alpha - 8)} \over 4}$

${{\alpha \pm (\alpha - 8)} \over 4} \Rightarrow {{2\alpha - 8} \over 4},2$

$ \Rightarrow {{\alpha - 4} \over 2},2$ (rejected)

To exists solutions

$ - 1 \le {{\alpha - 4} \over 2} \le 1 \Rightarrow - 2 \le \alpha - 4 \le 2 \Rightarrow 2 \le \alpha \le 6$

$ \therefore $ $\alpha \in \left[ {2,6} \right]$

if $\theta $ $ \in $ $\left( {{\pi \over 2},{{2\pi } \over 3}} \right)$ $ \Rightarrow \cos \theta \in \left( { - {1 \over 2},0} \right) \Rightarrow - \cos \theta \in \left( {0,{1 \over 2}} \right)$

$\sin \theta \in \left( {{{\sqrt 3 } \over 2},1} \right)$ and $\cot \theta \in \left( { - {1 \over {\sqrt 3 }},0} \right)$

If $\theta \in \left( {\pi ,{{7\pi } \over 6}} \right) \Rightarrow \cos \theta \in \left( { - 1,{{ - \sqrt 3 } \over 2}} \right) \Rightarrow - \cos \theta \in \left( {{{\sqrt 3 } \over 2},1} \right)$

$\sin \theta \in \left( { - {1 \over 2},0} \right)$ and $\cot \theta \in \left( {\sqrt 3 ,\infty } \right)$

Then in $\theta $ $ \in $ $\left( {{\pi \over 2},{{2\pi } \over 3}} \right)$ $ \Rightarrow $ [sin$\theta $] = 0; [-cos$\theta $] = 0; [cot$\theta $] = -1;

Hence 0 = 0 and -x + y = 0 [have infinitely solutions]

and in $\theta \in \left( {\pi ,{{7\pi } \over 6}} \right)$ $ \Rightarrow $ [sin$\theta $] = -1; [-cos$\theta $] = 0; [cot$\theta $] = 1, 2. 3 ...........;

Then -x - 0.y = 0 $ \Rightarrow $ x = 0

$[\cot \theta ]x + y = 0$

$ \Rightarrow $ 1.x + y = 0 or 2x + y = 0 or ....... each of the line will cut x = 0 at exactly one point.

Hence unique solutions.

$\mathop {\underline {1 + {{\sin }^4}x} }\limits_{ \ge 1} = \mathop {\underline {{{\cos }^2}3x} }\limits_{ \le 1} $

Now for equality to hold sin⁴x = 0 & cos²3x = 1

sin⁴ = 0 $ \Rightarrow $ x = -2$\pi $, - $\pi $, 0, $\pi $, 2$\pi $

All of which satisfy cos²3x = 1 $ \Rightarrow $ 5 solutions

2cos²$\theta $ + 3sin$\theta $ = 0

$ \Rightarrow $ 2(1 - sin²$\theta $) + 3sin$\theta $ = 0

$ \Rightarrow $ 2sin²$\theta $ - 3sin$\theta $ - 2 = 0

$ \Rightarrow $ 2sin²$\theta $ - 4sin$\theta $ + sin$\theta $ - 2 = 0

$ \Rightarrow $ (2sin$\theta $ + 1)(sin$\theta $ - 2) = 0

$ \therefore $ sin$\theta $ = 2 (Not possible)

sin$\theta $ = $ - {1 \over 2}$

$ \therefore $ $\theta $ = n$\pi $ + (-1)ⁿ$\left( { - {\pi \over 6}} \right)$

When n = 0, $\theta $ = ${ - {\pi \over 6}}$

When n = 1, $\theta $ = ${{7\pi } \over 6}$

When n = -1, $\theta $ = ${-{5\pi } \over 6}$

When n = 2, $\theta $ = ${{11\pi } \over 6}$

$ \therefore $ Sum of all solutions in [–2$\pi $, 2$\pi $] is

= $ - {\pi \over 6} + {{7\pi } \over 6} - {{5\pi } \over 6} + {{11\pi } \over 6}$ = 2$\pi $

sin²2$\theta $ + cos⁴2$\theta $ = ${3 \over 4}, $$\theta $ $ \in $ $\left( {0,{\pi \over 2}} \right)$

$ \Rightarrow $  1 $-$ cos²2$\theta $ + cos⁴2$\theta $ = ${3 \over 4}$

$ \Rightarrow $  4cos2$\theta $ $-$ 4cos²2$\theta $ + 1 = 0

$ \Rightarrow $  (2cos²2$\theta $ $-$ 1)² = 0

$ \Rightarrow $  cos²2$\theta $ = ${1 \over 2}$ = cos²${{\pi \over 4}}$

$ \Rightarrow $  2$\theta $ = n$\pi $ $ \pm $ ${\pi \over 4}$, n $ \in $ I

$ \Rightarrow $  $\theta $ = ${{n\pi } \over 2} \pm {\pi \over 8}$

$ \Rightarrow $  $\theta $ = ${\pi \over 8},{\pi \over 2} - {\pi \over 8}$

Sum of solutions ${\pi \over 2}$

sin x $-$ sin 2x + sin 3x = 0        $x \in \left[ {0,{\pi \over 2}} \right)$

$ \Rightarrow $  (sin3x + sinx) $-$ sin2x = 0

$ \Rightarrow $  2sin2x.cos2x $-$ sin2x = 0

$ \Rightarrow $  sin2x (2cosx $-$ 1) = 0

sin 2x = 0

x = 0

and cos x = ${1 \over 2}$

and x = ${\pi \over 3}$

two solutions

As we know,

$\cos \left( {x + y} \right)\cos \left( {x - y} \right) = {\cos ^2}x - {\sin ^2}y$

$\therefore$ ${\mathop{\rm cos}\nolimits} \left( {{\pi \over 6} + x} \right)cos\left( {{\pi \over 6} - x} \right) = {\cos ^2}\left( {{\pi \over 6}} \right) - {\sin ^2}x$

Given, $8\cos x\left( {\cos \left( {{\pi \over 6} + x} \right)\cos \left( {{\pi \over 6} - x} \right) - {1 \over 2}} \right) = 1$

$ \Rightarrow 8\cos x\left( {{{\cos }^2}{\pi \over 6} - {{\sin }^2}x - {1 \over 2}} \right) = 1$

$ \Rightarrow 8\cos x\left( {{3 \over 4} - {1 \over 2} - {{\sin }^2}x} \right) = 1$

$ \Rightarrow 8\cos x\left( {{3 \over 4} - {1 \over 2} - 1 + {{\cos }^2}x} \right) = 1$

$ \Rightarrow 8\cos x\left( {{{\cos }^2}x - {3 \over 4}} \right) = 1$

$ \Rightarrow 2\left( {4{{\cos }^3}x - 3\cos x} \right) = 1$

(Taking $4\cos x$ inside the bracket).

We know,

$4{\cos ^3}x - 3\cos x = \cos 3x$

$ \Rightarrow 2\cos 3x = 1$

$ \Rightarrow \cos 3x = {1 \over 2}$

$\therefore$ $\,\,\,$ $3x = 2n\pi \pm {\pi \over 3}$

So, $x = {{2n\pi } \over 3} \pm {\pi \over 9}$

At $n=0,$ $x = + {\pi \over 9}$ as $x \in \left[ {0,\pi } \right]$

At $n=1,$ $x = {{2\pi } \over 3} + {\pi \over 9}$

$\therefore$ $\,\,\,$ $x = {{2\pi } \over 3} + {\pi \over 9}$

and $x = {{2\pi } \over 3} - {\pi \over 9}$

At $n = 2,$ $x = {{4\pi } \over 3} \pm {\pi \over 9}$

Which does not belongs to $\left[ {0,\pi } \right]$

So, possible values of $x$ is $ = {\pi \over 9},{{2\pi } \over 3} + {\pi \over 9},{{2\pi } \over 3} - {\pi \over 3}$

Sum of the solutions $ = {\pi \over 9} + {{2\pi } \over 3} + {\pi \over 9} + {{2\pi } \over 3} - {\pi \over 9}$ v

$ = {{4\pi } \over 3} + {\pi \over 9}$ $ = {{13\,\pi } \over 9}$

According to the question,

$k\pi = {{13\pi } \over 9}$

$\therefore\,\,\,$ $k = {{13} \over 9}$

sin 3x = cos 2x

$ \Rightarrow $$\,\,\,$ 3 sin x $-$ 4 sin³ x = 1 $-$ 2 sin² x

$ \Rightarrow $$\,\,\,$ 4 sin³ x $-$ 2 sin² x $-$ 3 sin x + 1 = 0

$ \Rightarrow $$\,\,\,$ sin x = 1, ${{ - 2 \pm 2\sqrt 5 } \over 8}$

In the interval $\left( {{\pi \over 2},\pi } \right)$, sin x = ${{ - 2 + 2\sqrt 5 } \over 8}$

So, there is only one solution.

We have,

$ \left|\sqrt{2 \sin ^4 x+18 \cos ^2 x}-\sqrt{2 \cos ^4 x+18 \sin ^2 x}\right|=1 $

That is,

$ f(x)=\left|\sqrt{2 \sin ^4 x+8 \cos ^2 x}-\sqrt{2 \cos ^4 x+18 \sin ^2 x}\right| $

Now, $f(x)=f\left(\frac{\pi}{2}-x\right) $

$\Rightarrow f(x)$ is symmetric about $x=\frac{\pi}{4}$.

If $f(x)$ has solution in $(0, \pi / 4)$, then in $(0,2 \pi)$, there are eight solutions.

$\cos x + \cos 2x + \cos 3x + \cos 4x = 0$

$ \Rightarrow $ $(\cos x + \cos 3x)$ + $(\cos 2x + \cos 4x)$ = 0

$ \Rightarrow 2\cos 2x\cos x + 2\cos 3x\cos x = 0$

$ \Rightarrow 2\cos x\left( {2\cos {{5x} \over 2}\cos {x \over 2}} \right) = 0$

$\cos x = 0,\cos {{5x} \over 2} = 0,\cos {x \over 2} = 0$

$x = \pi ,{\pi \over 2},{{3\pi } \over 2},{\pi \over 5},{{3\pi } \over 5},{{7\pi } \over 5},{{9\pi } \over 5}$

$2{\sin ^2}x + 5\sin x - 3 = 0$

$ \Rightarrow \left( {\sin x + 3} \right)\left( {2\sin x - 1} \right) = 0$

$\sin x = {1 \over 2}$ and $\,\,\sin x \ne - 3$

Given that $x \in \left[ {0,3\pi } \right]$

So possible values of x are $30^\circ $, $150^\circ $, $390^\circ $, $510^\circ $. That means x have 4 values.

We can simplify the equation by converting everything to sines and cosines:

$\frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2\cos x$

Multiplying through by $\cos x$ gives:

$\sin x + 1 = 2\cos^2 x$

Using the identity $\cos^2 x + \sin^2 x = 1$, we can substitute $\sin^2 x$ with $1 - \cos^2 x$ to get:

$\sin x + 1 = 2 - 2\sin^2 x$

Rearranging terms gives:

$2\sin^2 x + \sin x - 1 = 0$

This is a quadratic equation in $\sin x$, which we can solve using the quadratic formula:

$\sin x = \frac{-1 \pm \sqrt{1 + 8}}{4}$

The discriminant is positive, so there are two solutions for $\sin x$:

$\sin x = \frac{-1 \pm \sqrt{9}}{4} = -1, \frac{1}{2}$

For $\sin x = -1$, we have $x = \frac{3\pi}{2}$.

For $\sin x = \frac{1}{2}$, we have $x = \frac{\pi}{6}, \frac{5\pi}{6}$.

Therefore, there are a total of $\boxed{3}$ solutions in the interval $\left[ 0, 2\pi \right]$.