The number of solution-s of 2 sin2 x - sin2 2x -2- x - - 0 -2 - - are -

(1 cos 2x ) +( 1 cos^2 2x )=2 ⇒cos 2x ( cos 2x +1 ) =0 ⇒ nbsp;cos 2x =0 or cos 2x = 1 ⇒ 2x= (2n+1) π2 or 2x= (2n 1) π , n ∈ℤ ⇒ x= (2n+1) π4 or nbsp;x= ...

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Standard XII

Mathematics

Solving Simultaneous Trigonometric Equations

The number of...

Question

The number of solution's of $2 {sin}^{2} x + {sin}^{2} 2 x = 2, x \in [0, 2 π]$ are :

4

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5

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7

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6

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Solution

The correct option is D $6$
$(1 - cos 2 x) + (1 - {cos}^{2} 2 x) = 2$
$\Rightarrow cos 2 x (cos 2 x + 1) = 0$
$\Rightarrow cos 2 x = 0$ or $cos 2 x = - 1$
$\Rightarrow 2 x = (2 n + 1) \frac{π}{2}$ or $2 x = (2 n - 1) π, n \in Z$
$\Rightarrow x = (2 n + 1) \frac{π}{4} or x = (2 n - 1) \frac{π}{2}, n \in Z$
Hence, the solutions are $\frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}, \frac{π}{2}, \frac{3 π}{2}$

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The number of solution's of 2sin2x+sin22x=2,x∈[0,2π] are :

The correct option is D 6(1−cos2x)+(1−cos22x)=2 ⇒cos2x(cos2x+1)=0 ⇒cos2x=0 or cos2x=−1 ⇒2x=(2n+1)π2 or 2x=(2n−1)π, n∈Z ⇒x=(2n+1)π4 or x=(2n−1)π2, n∈Z Hence, the solutions are π4,3π4,5π4,7π4,π2,3π2

The number of solution's of $2 {sin}^{2} x + {sin}^{2} 2 x = 2, x \in [0, 2 π]$ are :