Question

if 2cos²Θ+cosΘ-1=0 find Θ

Answer 1

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

Given Trigonometric equation is

$\rm :\longmapsto\: {2cos}^{2}\theta + cos\theta - 1 = 0$

Using splitting of middle terms, we get

$\rm :\longmapsto\: {2cos}^{2}\theta +2 cos\theta - cos\theta - 1 = 0$

$\rm :\longmapsto\:2cos\theta (cos\theta + 1) - 1(cos\theta + 1) = 0$

$\rm :\longmapsto\:(2cos\theta - 1)(cos\theta + 1) = 0$

$\rm :\longmapsto\:cos\theta = \dfrac{1}{2} \: \: or \: \: cos\theta = - 1$

Case :- 1

$\rm :\longmapsto\:cos\theta = \dfrac{1}{2}$

$\rm :\longmapsto\:cos\theta = cos\dfrac{\pi}{3}$

We know

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

So, using this identity, we get

$\rm \implies\:\boxed{ \tt{ \: \theta = 2n\pi \pm \: \frac{\pi}{3} \: \forall \: n \in \: Z \:}}$

Case :- 2

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

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

$\rm \implies\:\boxed{ \tt{ \: \theta = 2n\pi \pm \:\pi \: \forall \: n \in \: Z \:}}$

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Additional Information :-

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