Keerthana
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अगर 7sin2θ+3cos2θ=4(0θ90)7 \sin ^{2} \theta+3 \cos ^{2} \theta=4\left(0^{\circ} \leq \theta \leq 90^{\circ}\right), तो मूल्य

$\thete$ is

a

π2\frac{\pi}{2}

b

π3\frac{\pi}{3}

c

π6\frac{\pi}{6}

d

π4\frac{\pi}{4}

Answer : Option C
Explanation :

7sin2θ+3cos2θ=47 \sin ^{2} \theta+3 \cos ^{2} \theta=4

7sin2θ+3(1sin2θ)=4\Rightarrow 7 \sin ^{2} \theta+3\left(1-\sin ^{2} \theta\right)=4

7sin2θ+33sin2θ=4\Rightarrow 7 \sin ^{2} \theta+3-3 \sin ^{2} \theta=4

4sin2θ=43=1\Rightarrow 4 \sin ^{2} \theta=4-3=1

sin2θ=14sinθ=12=sinπ6θ=π6\Rightarrow \sin ^{2} \theta=\frac{1}{4} \Rightarrow \sin \theta=\frac{1}{2}=\sin \frac{\pi}{6} \Rightarrow \theta=\frac{\pi}{6}

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