Trigonometrical Equations 1 Question 30
30. Solve $2(\cos x+\cos 2 x)+(1+2 \cos x) \sin 2 x$
$$ =2 \sin x,-\pi \leq x \leq \pi $$
(1978, 3M)
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Solution:
- Given that,
$2 \cos x+2 \cos 2 x+\sin 2 x+\sin 3 x+\sin x-2 \sin x=0$
$\therefore \quad 2 \cos x+2 \cos 2 x+2 \sin x \cos x+(\sin 3 x-\sin x)=0$
$\Rightarrow \quad 2 \cos x+2 \cos 2 x+2 \sin x \cos x+2 \cos 2 x \sin x=0$
$\Rightarrow \quad 2 \cos x(1+\sin x)+2 \cos 2 x(1+\sin x)=0$
$\Rightarrow \quad 2(1+\sin x)(\cos x+\cos 2 x)=0$
$\Rightarrow \quad 4(1+\sin x) \cos \frac{3 x}{2} \cos \frac{x}{2}=0$
$\therefore \quad 1+\sin x=0$
or $\quad \cos \frac{3 x}{2}=0$ or $\cos \frac{x}{2}=0$
If $1+\sin x=0$, then $\sin x=-1$
$\therefore \quad x=2 n \pi+\frac{3 \pi}{2}$
If $\cos \frac{3 x}{2}=0$, then $\frac{3 x}{2}=(2 n+1) \frac{\pi}{2}$
$\therefore \quad x=(2 n+1) \frac{\pi}{3}$
And if $\cos \frac{x}{2}=0$, then $\frac{x}{2}=(2 n+1) \frac{\pi}{2}$
$\therefore \quad x=(2 n+1) \pi$
But given interval is $[-\pi, \pi]$.
Put $n=-1$ in Eq. (i), $x=-\frac{\pi}{2}$
Put $n=0,1,-1,-2$ in Eq. (ii), $x=\frac{\pi}{3}, \pi-\frac{\pi}{3},-\pi$
Hence, the solution in $[-\pi, \pi]$ are $-\pi,-\frac{\pi}{2},-\frac{\pi}{3}, \frac{\pi}{3}, \pi$.
