Question

plss don't guess answer​

Answer 1

Step-by-step explanation:

We have,

$3 \cos( \theta) - 4 \cos^{3} ( \theta) = 1$

$\implies3 \cos( \theta) - 4 \cos^{3} ( \theta) - 1 = 0$

$\implies 4 \cos^{3} ( \theta) - 3 \cos( \theta) + 1 = 0$

$\implies 4 \cos^{3} ( \theta) + 4 \cos^{2} ( \theta) - 4 \cos ^{2} ( \theta) - 4 \cos( \theta) + \cos( \theta) + 1 = 0$

$\implies 4 \cos^{2} ( \theta) \{ \cos( \theta) + 1 \} - 4 \cos ( \theta) \{ \cos( \theta) + 1 \} + \{ \cos( \theta) + 1 \}= 0$

$\implies \{ \cos( \theta) + 1 \} \{ 4 \cos^{2} ( \theta)- 4 \cos ( \theta) + 1 \}= 0$

$\implies \cos( \theta) + 1 = 0 \: \: \: or \: \: \: 4 \cos^{2} ( \theta)- 4 \cos ( \theta) + 1 = 0$

$\implies \cos( \theta) + 1 = 0 \: \: \: or \: \: \: \{ 2 \cos ( \theta)- 1 \}^{2} = 0$

$\implies \cos( \theta) = - 1 \: \: \: or \: \: \: \cos ( \theta) = \frac{1}{2}$

$\implies \cos( \theta) = \cos ( \pi) \: \: \: or \: \: \: \cos ( \theta) = \cos \bigg( \frac{\pi}{3} \bigg)$

$\implies \theta = (2n + 1) \pi \: \: \: or \: \: \: \theta = 2m\pi \pm \frac{\pi}{3}$

Where, $n,m\:\in\mathbb{N}$

Now, since, $\theta\in[-\pi,\pi]$, so,

the solution becomes

$\theta = \pm \pi, \pm \frac{\pi}{3} ,$

Answer 2

$\large\underline{\sf{Given \:Question - }}$

The number of solution(s) for the equation

$\rm :\longmapsto\:3cos\theta - 4 {cos}^{3}\theta = 1 \: in \: \: \theta \: \in \: [ - \pi, \: \pi]$

$\large\underline{\sf{Solution-}}$

Given Trigonometric equation is

$\rm :\longmapsto\:3cos\theta - 4 {cos}^{3}\theta = 1 \: in \: \: \theta \: \in \: [ - \pi, \: \pi]$

can be rewritten as

$\rm :\longmapsto\: - ( - 3cos\theta + 4 {cos}^{3}\theta )= 1$

$\rm :\longmapsto\: 4 {cos}^{3}\theta - 3cos\theta= - 1$

We know,

$\red{ \boxed{ \sf{ \:cos3x = {4cos}^{3}x - 3cosx}}}$

So, using this, we get

$\rm :\longmapsto\:cos3\theta = - 1$

$\rm :\longmapsto\:cos3\theta = cos\pi$

We know,

$\red{ \boxed{ \sf{ \:cosx = cosy \: \bf\implies \:x = 2n\pi \: \pm \: y \: \: \forall \: n \: \in \: Z}}}$

So, using this,

$\rm :\longmapsto\:3\theta = 2n\pi \: \pm \: \pi \: \: \forall \: n \: \in \: Z$

$\rm :\longmapsto\:3\theta = (2n\: \pm \: 1) \: \pi \: \: \forall \: n \: \in \: Z$

$\bf :\longmapsto\:\theta = (2n\: \pm \: 1) \:\dfrac{\pi}{3} \: \: \forall \: n \: \in \: Z$

So,

As it is given that,

$\rm :\longmapsto\: - \pi \leqslant \theta \leqslant \pi$

So,

$\bf\implies \:\theta = - \dfrac{\pi}{3}, \: \dfrac{\pi}{3}, \: \pi, \: - \: \pi$

So,

$\bf\implies \:\theta \: assumes \: 4 \: values \:$

Thus,

$\bf\implies \:Number \: of \: solutions \: = \: 4$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf T-eq & \bf Solution \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf sinx = 0 & \sf x = n\pi\: \forall \: n \: \in \: Z \\ \\ \sf cosx = 0 & \sf x = (2n + 1)\dfrac{\pi}{2}\: \forall \: n \: \in \: Z\\ \\ \sf tanx = 0\: & \sf x = n\pi\: \forall \: n \: \in \: Z\\ \\ \sf sinx = siny & \sf x = n\pi + {( - 1)}^{n}y \\ \\ \sf cosx = cosy & \sf x = 2n\pi \pm \: y\: \forall \: n \: \in \: Z\\ \\ \sf tanx = tany & \sf x = n\pi + y\: \forall \: n \: \in \: Z \end{array}} \\ \end{gathered}\end{gathered}$