Question

4 sin² θ - 3 = 0, when 0° < θ < 90°​

Answer 1

$\huge\boxed{\underline{\bf\green{A} \: \red{N} \: \orange{S} \: \purple{W} \: \blue{E} \: \pink{R}}}$

$\longrightarrow \sf4 { \sin}^{2} \theta - 3 = 0 \\ \\ \longrightarrow \sf4 { \sin}^{2} \theta = 3 \\ \\ \longrightarrow \sf{ \sin}^{2} \theta = \frac{3}{4} \\ \\ \longrightarrow \sf \sin \theta = \sqrt{ \frac{3}{4} } \\ \\ \longrightarrow \sf \sin \theta = \frac{ \sqrt{3} }{2} \\ \\ \longrightarrow \sf \sin \theta = \sin60 \degree \\ \\\longrightarrow \boxed{ \sf \theta = 60 \degree}$

Answer 2

Answer:

I hope it help

Step-by-step explanation:

It is a right answer