Solve for xR02pi 2cos2x 7cosx 4 0 Do I use u substitution le

Solve for xR[0,2pi]: 2cos^2(x) -7cos(x) -4 =0.

Do I use u substitution? let cos(x) = u

2u^2-7u-4=0; (2u+1)(u-4); u= -1/2, 4

cos(x)= -1/2

x= 2pi/3 + 2pik, element of the reals ER(backwards E)

cos(x)=4, is 4 to big of range, so do I ignore 4. I need someon to clarify my work and help me solve it the right way if I did it wrong. Thanks

Solution

you solved it correctly up to cos(x)=-1/2 there are two solutions 2pi/3 + 2pik and -2pi/3 + 2pik. we are only looking for solution in the interval [0,2*pi] so just 2*pi/3 and -2*pi/3+2*pi which equals 4*pi/3. so x=2*pi/3 and x=4*pi/3 are the solutions ignore cos(x)=4 because it is not possible.