The solution set of the equation $$4 \sin \theta-2 \cos \theta-2\sqrt {3} \sin \theta+\sqrt {3}=0$$ in the interval $$(0,2\pi)$$ is-
Explanation
Simplifying $$\sin 3x = \cos 2x$$.
$$\sin 3x = \cos 2x$$
$$3\sin x - 4{\sin ^3}x = 1 - 2{\sin ^2}x$$
$$4{\sin ^3}x - 2{\sin ^2}x - 3\sin x + 1 = 0$$
Put $$\sin x = t$$.
$$4{t^3} - 2{t^2} - 3t + 1 = 0$$
$$\left( {t - 1} \right)\left( {4{t^2} + 2t - 1} \right) = 0$$
$$t = 1$$
Or,
$$4{t^2} + 2t - 1 = 0$$
$$t = \frac{{ - 2 \pm \sqrt {{{\left( 2 \right)}^2} - 4\left( 4 \right)\left( { - 1} \right)} }}{{2 \times 4}}$$
$$ = \frac{{ - 2 \pm \sqrt {20} }}{8}$$
$$ = \frac{{ - 2 \pm 2\sqrt 5 }}{8}$$
$$ = \frac{{ - 1 \pm \sqrt 5 }}{4}$$
Then,
$$\sin x = 1$$
$$x = \frac{\pi }{2}$$
$$\sin x = \frac{{ - 1 \pm \sqrt 5 }}{4}$$
$$\sin x = \frac{{ - 1 + \sqrt 5 }}{4}$$
$$x = \pi + \frac{\pi }{{10}}$$
$$x = \frac{{11\pi }}{{10}}$$
$$\sin x = \frac{{ - 1 - \sqrt 5 }}{4}$$
$$x = 2\pi - \frac{\pi }{{10}}$$
$$x = \frac{{19\pi }}{{10}}$$
So, $$x = \frac{\pi }{2}$$, $$x = \frac{{11\pi }}{{10}}$$ and $$x = \frac{{19\pi }}{{10}}$$.
Since the given interval is $$\left( {\frac{\pi }{2},\pi } \right)$$, then, $$x = \frac{\pi }{2}$$ is the only solution.
Therefore, the number of solutions of $$\sin 3x = \cos 2x$$ is 1.
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