Trigonometrical Equations
hard

Let $S$ be the sum of all solutions (in radians) of the equation $\sin ^{4} \theta+\cos ^{4} \theta-\sin \theta \cos \theta=0$ in $[0,4 \pi]$ Then $\frac{8 \mathrm{~S}}{\pi}$ is equal to ...... .

A

$87$

B

$78$

C

$56$

D

$65$

(JEE MAIN-2021)

Solution

Given equation

$\sin ^{4} \theta+\cos ^{4} \theta-\sin \theta \cos \theta=0$

$\Rightarrow 1-\sin ^{2} \theta \cos ^{2} \theta-\sin \theta \cos \theta=0$

$\Rightarrow 2-(\sin 2 \theta)^{2}-\sin 2 \theta=0$

$\Rightarrow(\sin 2 \theta)^{2}+(\sin 2 \theta)-2=0$

$\Rightarrow(\sin 2 \theta+2)(\sin 2 \theta-1)=0$

$\Rightarrow \sin 2 \theta=1 \text { or } \sin 2 \theta=-2$ $\rightarrow$ not possible

$\Rightarrow \quad 2 \theta=\frac{\pi}{2}, \frac{5 \pi}{2}, \frac{9 \pi}{2}, \frac{13 \pi}{2}$

$\Rightarrow \quad \theta=\frac{\pi}{4}, \frac{5 \pi}{4}, \frac{9 \pi}{4}, \frac{13 \pi}{4}$

$\Rightarrow \frac{8 S}{\pi}=\frac{8 \times 7 \pi}{\pi}=56.00$

Standard 11
Mathematics

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