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Trigonometric Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsTrigonometric Functions2 Marks
MCQ
If $2 \sin ^2 x+\sin ^2 2 x=2,-\pi < x <\pi $, then $x=$
  • A
    $\pm \frac{\pi}{6}$
  • ✓
    $\pm \frac{\pi}{4}$
  • C
    $\frac{3 \pi}{2}$
  • D
    none of these
✓

Answer

Correct option: B.
$\pm \frac{\pi}{4}$
(B) $2 \sin ^2 x+\sin ^2 2 x=2$
$\Rightarrow(1-\cos 2 x)+\left(1-\cos ^2 2 x\right)=2$ $\ldots .\left[\because \sin ^2 \theta+\cos ^2 \theta=1\right.$ and $2 \sin ^2 \theta=1-\cos 2 \theta$ ]
$\Rightarrow \cos 2 x(\cos 2 x+1)=0$
$\Rightarrow \cos 2 x=0 \text { or } \cos 2 x=-1$
$\Rightarrow 2 x=(2 n+1) \frac{\pi}{2}$ or $(2 n+1) \pi$
$\Rightarrow x=(2 n+1) \frac{\pi}{4}$ or $(2 n+1) \frac{\pi}{2}$
Putting $n =-2,-1,0,1,2$, we get
$x=\frac{-3 \pi}{4}, \frac{-\pi}{4}, \frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}$
and $\frac{-3 \pi}{2}, \frac{-\pi}{2}, \frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}$
Since, $-\pi < x <\pi$
$\therefore x=±\frac{\pi}{4}, ±\frac{\pi}{2}, ±\frac{3 \pi}{4}$
$\therefore$ option (B) is the correct answer.

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