Question

General solution of ${\sin }^{3}x+{\cos }^{3}x+\frac{3}{2}\sin 2x=1$

• A. $x=n\pi$ where $n$ is even integer
• B. $x=n\pi +\frac{\pi }{2}$ where $n$ is odd integer
• C. $x=2n\pi$ where $n$ is odd integer
• D. $x=n\pi -\frac{\pi }{2}$ where $n$ is even integer

Answer 1

The correct option is C $x=2n\pi$ where $n$ is odd integer

${\sin }^{3}x+{\cos }^{3}x+\frac{3}{2}\sin 2x=1$

$\Rightarrow (\sin x+\cos x)({\sin }^{2}x+{\cos }^{2}x-\sin x\cos x)+\frac{3}{2}\times 2\sin x\cos x=1$

$\Rightarrow (\sin x+\cos x)({\sin }^{2}x+{\cos }^{2}x-\sin x\cos x)+3\sin x\cos x=1$ ....$\left(1\right)$

Take $\Rightarrow \sin x+\cos x=t$

Squaring on both sides

$\Rightarrow (\sin x+\cos x{)}^2=t^2$

$\Rightarrow {\sin }^2+{\cos }^2+2\sin x\cos x=t^2$

$\Rightarrow 1+2\sin x\cos x=t^2$

$\Rightarrow \sin x\cos x=\frac{t^2}{2}-\frac{1}{2}$

Substituting these values in equation $\left(1\right)$ we get

$\Rightarrow t(1+\frac{t^2}{2}-\frac{1}{2})+3(\frac{t^2}{2}-\frac{1}{2})$

$\Rightarrow t^3+3t^2+t-5=0$

$\Rightarrow (t-1)(t^2+4t+5)=0$

Therefore, $\Rightarrow \sin x+\cos x=1$

Squaring on both sides

$\Rightarrow (\sin x+\cos x{)}^2=1$

$\Rightarrow 1+\sin 2x=1$

$\Rightarrow 2\sin x\cos x=0$

$\Rightarrow \sin x=0;\cos x=0$

$\Rightarrow x=2n\pi$