Question

Pls help trig question!!

Answer 1

$4\sin^2 x - 3 = 3 \sin x\\\\\implies 4\sin^2 x -3 \sin x -3 =0\\\\\implies 4u^2 -3u -3 =0~~~~~~~~~~~~~~;[\text{Set} ~ \sin x = u]\\\\\implies u = \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4\cdot 4 \cdot (-3)}}{2(4)}~~~~~~;[\text{Apply quadratic formula}]\\\\\\\implies u = \dfrac{3\pm\sqrt{57}}8\\\\\implies \sin x = \dfrac{3\pm\sqrt{57}}8~~~~~~;[\text{Substitute back}~ u = \sin x]\\\\\text{Now,}\\\\\sin x = \dfrac{3+\sqrt{57}}8\\\\\text{No solution for}~ x\in \mathbb{R} ~\text{Since}~ -1\leq \sin x \leq 1$

$\sin x = \dfrac{3 - \sqrt{57}}8\\\\\implies x = n\pi + (-1)^n \sin^{-1} \left(\dfrac{3-\sqrt{57}}8\right)\\\\\\\text{ When n = 1,2 and}~ [0,2\pi]},\\\\\\x= \pi - \sin^{-1} \left(\dfrac{3-\sqrt{57}}8 \right)~ \text{and} ~x = 2\pi + \sin^{-1} \left(\dfrac{3-\sqrt{57}}8 \right)\\\\\text{In decimal,}\\\\x= 3.75~~~~ \text{and}~~~ x =5.68$