If 2 sin θ = √3, cos θ = ?

Given:

2sinθ=3–√2sin⁡θ=3 $2 \sin \theta = \sqrt{3}$

Formula used:

We know the Pythagorean identity: sin2θ+cos2θ=1sin2⁡θ+cos2⁡θ=1 $\sin^2 \theta + \cos^2 \theta = 1$

Calculation:

First, solve for sinθsin⁡θ $\sin \theta$ :

2sinθ=3–√2sin⁡θ=3 $2 \sin \theta = \sqrt{3}$

sinθ=3√2sin⁡θ=32 $\sin \theta = \frac{\sqrt{3}}{2}$

Using the Pythagorean identity:

sin2θ+cos2θ=1sin2⁡θ+cos2⁡θ=1 $\sin^2 \theta + \cos^2 \theta = 1$

(3√2)2+cos2θ=1(32)2+cos2⁡θ=1 $\left( \frac{\sqrt{3}}{2} \right)^2 + \cos^2 \theta = 1$

34+cos2θ=134+cos2⁡θ=1 $\frac{3}{4} + \cos^2 \theta = 1$

cos2θ=1−34cos2⁡θ=1−34 $\cos^2 \theta = 1 - \frac{3}{4}$

cos2θ=14cos2⁡θ=14 $\cos^2 \theta = \frac{1}{4}$

cosθ=±12cos⁡θ=±12 $\cos \theta = \pm \frac{1}{2}$

Therefore, the value of cosθcos⁡θ $\cos \theta$ is ±12±12 $\pm \frac{1}{2}$ .

So, the correct option is:

2) 1212 $\frac{1}{2}$