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Trigonometrical Equations
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The total number of solution of $sin^4x + cos^4x = sinx\, cosx$ in $[0, 2\pi ]$ is equal to
A
$2$
B
$4$
C
$6$
D
none of these
Solution
$\sin ^{4} x+\cos ^{4} x=\sin x \cos x$
or $\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x=\sin x \cos x$
or $1-\frac{\sin ^{2} 2 x}{2}=\frac{\sin 2 x}{2}$
or $\sin ^{2} 2 x+\sin 2 x-2=0$
$\operatorname{or}(\sin 2 x+2)(\sin 2 x-1)=0$
or $\sin 2 x=1$
or $2 \mathrm{x}=(4 \mathrm{n}+1) \frac{\pi}{2}, \mathrm{n} \in \mathrm{Z}$
or $x=(4 n+1) \frac{\pi}{4}, n \in Z$
$=\frac{\pi}{4}, \frac{5 \pi}{4} \quad(\because x \in[0,2 \pi])$
Thus, there are two solutions.
Standard 11
Mathematics
