Answer to Question #212421 in Trigonometry for Teba

Question

1. Determine the general solution of the equation: 4sin² x - 3 = 0

Answer 1

$4\sin^2x-3=0\\ 4\sin^2x=3\\ \sin^2x=\frac{3}{4}\\ \sin{x}=\pm{\sqrt{\frac{3}{4}}}\\ \sin{x}=\pm\frac{\sqrt{3}}{2}\\ x=\sin^{-1}{(\frac{\sqrt{3}}{2})}\\ x=60\degree, 120\degree\\ x=\sin^{-1}{(-\frac{\sqrt{3}}{2})}\\ x=300\degree, 240\degree\\$

General solution is

$x=360\degree(n)+60\degree, 360\degree(n) +120\degree, 360\degree(n)+240\degree, 360\degree(n)+300\degree$