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Trigonometrical Equations
hard
The number of solutions of the equation $\sin \theta+\cos \theta=\sin 2 \theta$ in the interval $[-\pi, \pi]$ is
A
$1$
B
$2$
C
$3$
D
$4$
(KVPY-2017)
Solution
(b)
We have, $\sin \theta+\cos \theta=\sin 2 \theta$ On squaring both sides, we get $\sin ^2 \theta+\cos ^2 \theta+\sin 2 \theta=\sin ^2 2 \theta$ $\Rightarrow \sin ^2 2 \theta-\sin 2 \theta-1=0$
$\Rightarrow \quad \sin 2 \theta=\frac{1 \pm \sqrt{5}}{2}$
$\Rightarrow \sin 2 \theta=\frac{1-\sqrt{5}}{2} \Rightarrow \sin 2 \theta \neq \frac{\sqrt{5}+1}{2}$
$\therefore$ Two solutions exist interval $[-\pi, \pi]$.
Standard 11
Mathematics
