Question

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

• A. $0$
• B. $\pi$
• C. $2\pi$
• D. $\frac{5\pi }{2}$

Answer 1

Answer: (C)

$2\pi$

$1+{\sin }^{4}x={\cos }^{2}3x$,

$1-{\cos }^{2}3x=-{\sin }^{4}x$

${\sin }^{2}3x=-{\sin }^{4}x$

$(3\sin x-4{\sin }^{3}x{)}^2=-{\sin }^{4}x$

${\sin }^{2}x(16{\sin }^{4}x-23{\sin }^{2}x+9)=0$

$\sin x=0$ or $16{\sin }^{4}x-23{\sin }^{2}x+9=0$ .....(D<0) (no real roots)

So, $\sin x=0$

$\Rightarrow x=-2\pi ,-\pi ,0,\pi ,2\pi$ when $x\in [-\frac{5\pi }{2},\frac{5\pi }{2}]$

But the greatest positive solution is $x=2\pi$

Hence option ${}^{\prime }{C}^{\prime }$ is the answer.