Question

Find all the solutions of $4{\cos }^{2}x\sin x-2{\sin }^{2}x=3\sin x$.

Answer 1

Find all the solutions of given equation

We can rewrite the given expression as

$\Rightarrow 4(1-{\sin }^{2}x)\sin x-2{\sin }^{2}x=3\sin x\mathbf{}\left[\because co{s}^{2}x+si{n}^{2}x=1\right]$

$\Rightarrow 4\sin x-4{\sin }^{3}x-2{\sin }^{2}x-3\sin x=0$

$\Rightarrow -4{\sin }^{3}x-2{\sin }^{2}x+\sin x=0$

$\Rightarrow -\sin x(4{\sin }^{2}x+2\sin x-1)=0$

$\Rightarrow \sin x=0$ or $\Rightarrow 4{\sin }^{2}x+2\sin x-1=0$

$\Rightarrow \sin x=\sin 0$ or $\Rightarrow \sin x=\frac{-2\pm \sqrt{4+16}}{2\left(4\right)}$

$\Rightarrow x=n\pi$ or $\Rightarrow \sin x=\frac{-1\pm \sqrt{5}}{4}$

$\Rightarrow x=n\pi$ or $\Rightarrow \sin x=\sin \frac{\pi }{10}$or $\Rightarrow \sin x=\sin \left(-\frac{3\pi }{10}\right)$

$\left[\because sin\left(\frac{\pi }{10}\right)=\frac{\sqrt{5}-1}{4},sin\left(-\frac{3\pi }{10}\right)=-\left(\frac{\sqrt{5}+1}{4}\right)\right]$

$\Rightarrow x=n\pi ,n\pi +{(-1)}^n\frac{\pi }{10},n\pi +{(-1)}^n\left(\frac{-3\pi }{10}\right)$ $\left[\because \mathrm{sin\theta }=\mathrm{sin\alpha }\Rightarrow \theta =\mathrm{n\pi }+{(-1)}^n\alpha \right]$

Hence, the general solution set is

$\mathbf{\{}\mathbf{x}\mathbf{:}\mathbf{x}\mathbf{=}\mathbf{n}\mathbf{\pi }\mathbf{\}}\mathbf{}\mathbf{\cup }\mathbf{}\mathbf{\{}\mathbf{x}\mathbf{:}\mathbf{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{n}\mathbf{\pi }\mathbf{}\mathbf{+}\mathbf{}{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{n}}\frac{\mathbf{\pi }}{\mathbf{10}}\mathbf{\}}\mathbf{\cup }\mathbf{\{}\mathbf{x}\mathbf{:}\mathbf{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{n}\mathbf{\pi }\mathbf{}\mathbf{+}\mathbf{}{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{n}}\mathbf{(}\mathbf{-}\frac{\mathbf{3}\mathbf{\pi }}{\mathbf{10}}\mathbf{)}\mathbf{\}}$