Question

Solution of the equation $4{\sin }^{4}x+{\cos }^{4}x=1$ is

• A. $x=n\pi$
• B. $x=2n\pi \pm {\cos }^{-1}\left(\sqrt{\frac{3}{5}}\right)$
• C. $x=(2n+1)\frac{\pi }{2}$
• D. none of these.

Answer 1

Answer: (B)

$4{\sin }^{4}x+{\cos }^{4}x=1$
$\Rightarrow$ $4(1-{\cos }^{2}x{)}^2+{\cos }^{4}x=1$
$\Rightarrow$ $4(1+{\cos }^{4}x-2{\cos }^{2}x)+co{s}^{4}x-1=0$
$\Rightarrow$ $5{\cos }^{4}x-8{\cos }^{2}x+3=0$
$\Rightarrow$ $(5{\cos }^{2}x-3)({\cos }^{2}x-1)=0$
If ${\cos }^{2}x$ = 1, then $x=n\pi ,n\in I$
If ${\cos }^{2}x=\frac{3}{2}$ , then $\cos x=\pm \sqrt{\frac{3}{5}}$
$\therefore$ $x=2n\pi \pm co{s}^{-1}\sqrt{\frac{3}{5}}$
Hence $x=2n\pi \pm {\cos }^{-1}\sqrt{\frac{3}{5}}$ gives the general value.