Question

19. Solve the equation 3cos x = 8tan x, between Oº and 360°

Answer 1

$3\cos x=5\tan x\\\\3\cos x=5\cdot\dfrac{\sin x}{\cos x}\\\\3\cos^2 x=5\sin x\\\\3(1-\sin^2x)=5\sin x\\\\-3\sin^2x-5\sin x+3=0\\\\t=\sin x\qquad\qqaud t\in\\\\-3t^2-5t+3=0\\\\3t^2+5t-3=0\quad\implies\quad a=3\,,\ b=5\,,\ c=-3\\\\t=\dfrac{-5\pm\sqrt{5^2-4\cdot3\cdot(-3)}}{2\cdot3}=\dfrac{-5\pm\sqrt{25+36}}{6}=\dfrac{-5\pm\sqrt{61}}{6}\\\\t_1=\dfrac{-5+\sqrt{61}}{6}\approx0.4684\ ,\qquad t_2=\dfrac{-5-\sqrt{61}}{6}\approx-2.136\notin$

$\sin x=0.4684\quad\wedge\quad x\in(0^o,\ 360^o)\\\\x\approx28^o\qquad\qquad\vee\qquad x\approx152^o\\\\\\\sin x=0.4684\quad\wedge\quad x\in(0,\ 2\pi)\\\\x\approx0.4887\qquad\qquad\vee\qquad x\approx2.6529$