Trigonometrical Equations 1 Question 29
29. Find all the solutions of $4 \cos ^{2} x \sin x-2 \sin ^{2} x=3 \sin x$.
$(1983,2 M)$
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Answer:
Correct Answer: 29. $x=-\pi,-\frac{\pi}{2},-\frac{\pi}{3}, \frac{\pi}{3}, \pi$
Solution:
- Given, $4 \cos ^{2} x \sin x-2 \sin ^{2} x=3 \sin x$
$\Rightarrow 4\left(1-\sin ^{2} x\right) \sin x-2 \sin ^{2} x-3 \sin x=0$
$\Rightarrow \quad 4 \sin x-4 \sin ^{3} x-2 \sin ^{2} x-3 \sin x=0$
$\Rightarrow \quad-4 \sin ^{3} x-2 \sin ^{2} x+\sin x=0$
$\Rightarrow \quad-\sin x\left(4 \sin ^{2} x+2 \sin x-1\right)=0$
$\Rightarrow \sin x=0 \quad$ or $\quad 4 \sin ^{2} x+2 \sin x-1=0$
$\Rightarrow \sin x=\sin 0 \quad$ or $\quad \sin x=\frac{-2 \pm \sqrt{4+16}}{2(4)}$
$\Rightarrow \quad x=n \pi \quad$ or $\quad \sin x=\frac{-1 \pm \sqrt{5}}{4}$
$\Rightarrow x=n \pi$ or $\sin x=\sin \frac{\pi}{10}$
or $\sin x=\sin -\frac{3 \pi}{10}$
$x=n \pi, n \pi+(-1)^{n} \frac{\pi}{10}, n \pi+(-1)^{n} \frac{-3 \pi}{10}$
$\therefore$ General solution set is
$$ \begin{aligned} {x: x=n \pi} & \cup x: x=n \pi+(-1)^{n} \frac{\pi}{10} \\ & \cup \quad x: x=n \pi+(-1)^{n} \frac{-3 \pi}{10} \end{aligned} $$
