Question

1. 4sin^2x=3tan^2x-1

2. 8cos^2x-4cos^4x=3

3. 3secx-cosx=2

Answer 1

Solve for values of x where 0 less/equal to x or less than 2pi. Express all answers in solution set form.

Answer 2

1.
4sin^2x = 3sin^2x/cos^2x - 1
multiply by cos^2x
4sin^2xcos^2x = 3sin^2x - cos^2x
4sin^2xcos^2x = 3sin^2 - (1 - sin^2x)
4sin^2xcos^2x = 4sin^2 - 1
4sin^2xcos^2x - 4sin^2 = - 1
4sin^2x(cos^2 - 1) = -1
4sin^2x(-sin^2x) = -1
-4sin^4x = -1
sin^4x = 1/4
sin^2 = ± 1/2
sinx = ±1/√2
x = π/4 , 3π/4 , 5π/4, and 7π/4

2.
let cos^2x = y, then we have
8y - 4y^2 = 3
4y^2 - 8y + 3 = 0
(2y - 1)(2y - 3) = 0
y = 1/2 or y = 3/2
so cos^2x = 1/2 or cos^2x = 3/2
cosx = ±1/√2 or cosx = ±√(3/2)
the second part is not possible since -1≤cosx≤+1
so cosx = ±1/√2
x = π/4 , 3π/4 , 5π/4, and 7π/4

try the third yourself, it is quite easy.