Question

The general solution of the equation ${\sin }^3\frac{x}{2}-{\cos }^3\frac{x}{2}=\frac{\cos x\times (2+\mathrm{sinx})}{3}$ is

• A. $2\mathrm{n\pi }-\frac{\pi }{2},n\in Z.$
• B. $2\mathrm{n\pi }+\frac{\pi }{2},n\in Z.$
• C. $2\mathrm{n\pi }-\frac{\pi }{4},n\in Z.$
• D. $2\mathrm{n\pi }+\frac{\pi }{4},n\in Z.$

Answer 1

The correct option is B $2\mathrm{n\pi }+\frac{\pi }{2},n\in Z.$

$\mathrm{We}\mathrm{are}\mathrm{given}\mathrm{that}{\sin }^3\frac{x}{2}-{\cos }^3\frac{x}{2}=\frac{\cos x\times \left(2+\mathrm{sinx}\right)}{3},\mathrm{Now},\mathrm{LHS}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as}{\sin }^3\frac{x}{2}-{\cos }^3\frac{x}{2}=\left(\sin \frac{x}{2}-\cos \frac{x}{2}\right)\left({\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}+\sin \frac{x}{2}\cos \frac{x}{2}\right)=\left(\sin \frac{x}{2}-\cos \frac{x}{2}\right)\frac{\left(}{2+\mathrm{sinx}}\mathrm{So},\mathrm{the}\mathrm{equation}\mathrm{becomes}\left(\sin \frac{x}{2}-\cos \frac{x}{2}\right)\frac{\left(}{2+\mathrm{sinx}}=\frac{\cos x\times \left(2+\mathrm{sinx}\right)}{3}\sin \frac{x}{2}-\cos \frac{x}{2}=\frac{2\cos x}{3}\sin \frac{x}{2}-\cos \frac{x}{2}=\frac{2\left({\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}\right)}{3}(\because \cos 2\theta ={\cos }^2\theta -{\sin }^2\theta )\sin \frac{x}{2}-\cos \frac{x}{2}=\frac{2\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)}{3}\sin \frac{x}{2}-\cos \frac{x}{2}=0\mathrm{or}\cos \frac{x}{2}+\sin \frac{x}{2}=\frac{-3}{2}\tan \frac{x}{2}=1\mathrm{or}\cos \frac{x}{2}+\sin \frac{x}{2}=\frac{-3}{2}\mathrm{Now},\tan \frac{x}{2}=1\Rightarrow \frac{x}{2}=\mathrm{n\pi }+\frac{\pi }{4},n\in Z\Rightarrow x=2\mathrm{n\pi }+\frac{\pi }{2},n\in Z\mathrm{or}\cos \frac{x}{2}+\sin \frac{x}{2}=\frac{-3}{2},\mathrm{this}\mathrm{equation}\mathrm{doesn}\text{'}t\mathrm{have}a\mathrm{solution}.\mathrm{If}\mathrm{asinx}+\mathrm{bcosx}=c\mathrm{have}a\mathrm{real}\mathrm{solution},\mathrm{then}\left(c\right|\le \sqrt{a^2+b^2}\mathrm{Hence},x=2\mathrm{n\pi }+\frac{\pi }{2},n\in Z$