The number of solutions of 1-sin xsin2 x2 - 0- where x - -0- 4- is

1+sin xsin^2x2=0 ⇒ 1+sin x(1 cos x)2 = 0 ⇒ 1+sin x2 sin 2 x4 = 0 ⇒sin2x=2sin x +4 As sin x ∈ [ 1, 1] ⇒ 2sin x +4 ∈ [2, 6] But sin 2x ∈ [ 1, 1] Hence, there is ...

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Standard XII

Mathematics

General Solution of Trigonometric Equation

The number of...

Question

The number of solution(s) of $1 + sin x {sin}^{2} \frac{x}{2} = 0$ , where $x \in [0, 4 π]$ , is

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Solution

The correct option is A $0$
$1 + sin x {sin}^{2} \frac{x}{2} = 0 \Rightarrow 1 + sin x \frac{(1 - cos x)}{2} = 0 \Rightarrow 1 + \frac{sin x}{2} - \frac{sin 2 x}{4} = 0 \Rightarrow sin 2 x = 2 sin x + 4$
As $sin x \in [- 1, 1]$
$\Rightarrow 2 sin x + 4 \in [2, 6]$
But $sin 2 x \in [- 1, 1]$

Hence, there is no solution.

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The number of solution(s) of 1+sinxsin2x2=0, where x∈[0,4π], is

The correct option is A 01+sinxsin2x2=0⇒1+sinx(1−cosx)2=0⇒1+sinx2−sin2x4=0⇒sin2x=2sinx+4 As sinx∈[−1,1] ⇒2sinx+4∈[2,6] But sin2x∈[−1,1] Hence, there is no solution.

The number of solution(s) of $1 + sin x {sin}^{2} \frac{x}{2} = 0$ , where $x \in [0, 4 π]$ , is