Question

If $xϵ[-\frac{5\pi }{2},\frac{5\pi }{2}]$, the greatest positive solution of $1+{\sin }^{4}x={\cos }^{2}3x$ is

• A. $\pi$
• B. $2\pi$
• C. $\frac{5\pi }{2}$
• D. none of these

Answer 1

Answer: (B)

$2\pi$

$1+si{n}^{4}x=co{s}^{2}3x$
$\Rightarrow 1+si{n}^{4}x=1-si{n}^{2}3x\Rightarrow si{n}^{4}x=-si{n}^{2}3x$
$\Rightarrow si{n}^{4}x=-(3sinx-4si{n}^{3}x{)}^2$
$\Rightarrow si{n}^{2}x(si{n}^{2}x+(3-4si{n}^{2}x{)}^2)=0$
$\Rightarrow si{n}^{2}x=0\Rightarrow sinx=0$
Hence solution in the given interval is,
$x=0,2\pi$
Thus greatest solution is $2\pi$.
Hence, option 'B' correct.