SOLUTION: Solve the following: 0 less then or equal to x less then 2pi A) 4cos x + sqrt 3= 2cos x B) 2sin^2 x - 5 sin x= -2 Thanks again for all the help!

Algebra ->  Trigonometry-basics -> SOLUTION: Solve the following: 0 less then or equal to x less then 2pi A) 4cos x + sqrt 3= 2cos x B) 2sin^2 x - 5 sin x= -2 Thanks again for all the help!      Log On


   

Question 744772: Solve the following: 0 less then or equal to x less then 2pi
A) 4cos x + sqrt 3= 2cos x

B) 2sin^2 x - 5 sin x= -2
Thanks again for all the help!
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
A) 4cos x + sqrt 3= 2cos x
4cos x - 2cos x = -sqrt(3)
2cosx=-sqrt(3)
cosx= -sqrt(3)/2
x= 7pi/6


B) 2sin^2 x - 5 sin x= -2
2sin^2 x - 5 sin x +2=0
2sin^2 x - 4 sin x -sinx+2=0
2sinx(sinx-2)-1(sinx-2)=0
(sinx-2)(2sinx-1)=0
sinx = 2
or
sinx =1/2
between 0 & 2pi find x