Trig Question, Please Help If You Know How To Do The Problem

Question

I keep having to ask for help with this lesson, so if you have time time and are able to help me, step-by-step would be a huge help.

Find x where pi ≤ x ≤ 2pi.

4 cos^2 x - 2 = sec^2 x - tan^2 x

Answer 1

x=7π/6, 11π/6 is correct

Answer 2

Simplify the right side of the equations to all sines and cosines.

4cos²x - 2 = 1/cos²x-sin²x/cos²x Multiply both sides of the equation by cos²x to eliminate the fractions

cos²x(4cos²x - 2) = cos²x(1/cos²x-sin²x/cos²x) Simplify

4cos⁴x-2cos²x=1-sin²x Use Pythagorean identity substitution to get rid of the sin on the right side

4cos⁴x-2cos²x=1-(1-cos²x) Simplify using the distributive property

4cos⁴x-2cos²x=1-1+cos²x Combine like terms

4cos⁴x-2cos²x-cos²x=cos²x cos²x Subtract cos²x from both sides so it equals zero

4cos⁴x-3cos²x=0 Factor out the GCF of cos²x

cos²x(4cos²x-3)=0 Set each factor equal to zero and solve

cos²x=0 Take the square root of both sides

cos x = 0

Between the values in the problem given the only solution to this is

x=3π/2

4cos²x-3=0 Add 3 to both sides

4cos²x-3+3=0+3 Simplify

4cos²x=3 Divide both sides by 4

4cos²x/4=3/4 Simplify

cos²x=3/4 Take the square root of both sides

cos x = ±√3/2

Between the values in the problem given the only solution to this is

x=7π/6, 11π/6

This is a very complicated problem. If you keep having to ask for help on them, you should probably meet one on one with a tutor for assistance.