Question

The total number of solution of ${\sin }^{4}x+{\cos }^{4}x=\sin x\cos x$ in $[0,2\pi )$ is equal to:

• A. $2$
• B. $4$
• C. $6$
• D. $None$

Answer 1

The correct option is A $2$

$si{n}^{4}x+co{s}^{4}x=sinx.cosx$

$\Rightarrow (si{n}^{2}x+co{s}^{2}x{)}^2-2si{n}^{2}x.co{s}^{2}x=sinx.cosx$

$\Rightarrow 1-2si{n}^{2}x.co{s}^{2}x=sinx.cosx$ $\Rightarrow 2(1-2si{n}^{2}x.co{s}^{2}x)=2sinx.cosx$

$\Rightarrow si{n}^{2}2x+sin2x-2=0\Rightarrow (sin2x-1)(sin2x+2)=0$

but $sin2x+2$ can't be zero, so $sin2x=1$ is only solution

therefore solutions in given interval are,

$x=\frac{\pi }{4},\frac{5\pi }{4}$

Hence, option 'A' is correct.