Question

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

O [A] 3.99,5.44
[B] 19.5, 160.5
[C] 0.197,3.34
O [D] 1.89, 5.03

Answer 1

`3\cos x=5\tan x\\\\3\cos x=5\cdot(\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<-1, 1>\\\\-3t^2-5t+3=0\\\\3t^2+5t-3=0\quad\implies\quad a=3\,,\ b=5\,,\ c=-3\\\\t=(-5\pm√(5^2-4\cdot3\cdot(-3)))/(2\cdot3)=(-5\pm√(25+36))/(6)=(-5\pm√(61))/(6)\\\\t_1=(-5+√(61))/(6)\approx0.4684\ ,\qquad t_2=(-5-√(61))/(6)\approx-2.136\\otin<-1,\,1>`

`\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`