Question

If 2 sin 2θ = $\sqrt{3}$, find the value of θ.
Or, If 7 sin²θ + 3 cos²θ = 4, show that tan θ = $\frac{1}{\sqrt{3}}$.

Answer 1

$$\begin{gathered} 2\sin 2\theta =\sqrt{3} \\ \Rightarrow \sin 2\theta =\frac{\sqrt{3}}{2} \\ \Rightarrow \sin 2\theta =\sin 60° \end{gathered}$$

$$\begin{gathered} \Rightarrow 2\theta =60° \\ \Rightarrow \theta =30° \end{gathered}$$

OR
7 sin²θ + 3 cos²θ = 4
⇒ 4 sin²θ + 3 sin²θ + 3 cos²θ = 4
⇒ 4 sin²θ + 3(sin²θ + cos²θ) = 4
⇒ 4 sin²θ + 3 = 4 [ since sin²θ + cos²θ = 1]
⇒ 4 sin²θ = 1
$\Rightarrow {\sin }^2\theta =\frac{1}{4}$
∴ ${\cos }^2\theta =\left(1-{\sin }^2\theta \right)=\left(1-\frac{1}{4}\right)=\frac{3}{4}$
∴ ${\tan }^2\theta =\frac{{\sin }^2\theta }{{\cos }^2\theta }=\left(\frac{1\times 4}{4\times 3}\right)=\frac{1}{3}$
$\Rightarrow \tan \theta =\frac{1}{\sqrt{3}}$