$\left\{x \in R: \cos 2 x+2 \cos ^2 x=2\right\}$ is equal to
$\left\{x \in R: \cos 2 x+2 \cos ^2 x=2\right\}$ is equal to
- $\left\{2 n \pi+\frac{\pi}{3}: n \in Z\right\}$
- $\left\{n \pi \pm \frac{\pi}{6}: n \in Z\right\}$
- $\left\{n \pi+\frac{\pi}{3}: n \in Z\right\}$
- $\left\{2 n \pi-\frac{\pi}{3}: n \in Z\right\}$
Solution
Given equation is
$
\begin{array}{rlrl}
& & \cos 2 x+2 \cos ^2 x & =2 \\
\Rightarrow & 2 \cos ^2 x-1+2 \cos ^2 x & =2 \\
\Rightarrow & 4 \cos ^2 x & =3 \\
\Rightarrow & \cos ^2 x & =\frac{3}{4} \\
\Rightarrow & & \cos x= \pm \frac{\sqrt{3}}{2} \\
& \therefore & x=n \pi \pm \frac{\pi}{6}: n \in Z
\end{array}
$
Asked in: AP EAMCET 2008
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