Damn instructors

Okay I have a horrible instructor. I need help with these practice problems

For trigonometry

"Find the general solutions of the eqasions algebraically"
(^2 is squared)
a. ( 4sin^2 x = 2cosx+1
b. sec(4x) =2

Thanks for any help.

>>2
rofl

>>1
Ok, for the first one replace 4*sin^2(x) with 4*(1 - cos^2(x)), then you get 4 - 4cos^2(x) = 2cos(x) + 1, or 4cos^2(x) + 2cos(x) - 3 = 0. Let u = cos(x), then this is 4u^2 + 2u - 3 = 0. Solve for u using the quadratic formula, and then use cos^-1 on those solutions to get solutions for x.

Second one: sec(4x) = 2, so 1/cos(4x) = 2, so cos(4x) = 1/2, so 4x = (pi/3) + 2n*pi or (-pi/3) + 2n*pi, and x = (pi/12) + n/2 * pi or (-pi/12) + n/2 * pi.
(n is any integer)