Question

Solve for$x:{\sin }^{4}x+{\cos }^{4}x=\frac{7}{2}\sin x.\cos x$

• A. $\frac{\mathrm{n\pi }}{2}+(-1{)}^n\frac{\pi }{12},n\in Z$
• B. $\frac{\mathrm{n\pi }}{2}+(-1{)}^n\frac{\pi }{6},n\in Z$
• C. $\mathrm{n\pi }+(-1{)}^n\frac{\pi }{6},n\in Z$
• D. $\mathrm{n\pi }+(-1{)}^n\frac{\pi }{12},n\in Z$

Answer 1

The correct option is A $\frac{\mathrm{n\pi }}{2}+(-1{)}^n\frac{\pi }{12},n\in Z$

Given: ${\sin }^{4}x+{\cos }^{4}x=\frac{7}{2}\sin x.\cos x$
$\Rightarrow ({\sin }^{2}x+{\cos }^{2}x{)}^2-2{\sin }^{2}x.{\cos }^{2}x=\frac{7}{2}\sin x.\cos x$
$\Rightarrow (1{)}^2-\frac{(2\sin x.\cos x{)}^2}{2}=\frac{7}{4}(2\sin x.\cos x\left)\right[\because {\sin }^{2}x+{\cos }^{2}x=1]$

$\Rightarrow (1{)}^2-\frac{(\sin 2x{)}^2}{2}=\frac{7}{4}\sin 2x[\because 2\sin x.\cos x=\sin 2x]$

$\Rightarrow 2{\sin }^{2}2x+7\sin 2x-4=0$
$\Rightarrow 2{\sin }^{2}2x+8\sin 2x-\sin 2x-4=0$
$\Rightarrow (2\sin 2x-1)(\sin 2x+4)=0$

$\Rightarrow \sin 2x=\frac{1}{2}\text{or}\sin 2x=-4$

$\therefore \sin 2x=\frac{1}{2}[\text{as}\sin 2x=-4\text{is not possible because}-1\le \sin \theta \le 1]$

$\Rightarrow 2x=n\pi +(-1{)}^n\frac{\pi }{6},n\in Z[\because \sin \theta =\sin \alpha \Rightarrow \theta =n\pi +(-1{)}^n\alpha ]$

$\Rightarrow x=\frac{n\pi }{2}+(-1{)}^n\frac{\pi }{12},n\in Z$