im working on my trig homework and i am stuck.?

Question

I have to sove for the angle in radians: 2 cos^2 x - cos x = 2 - sec x?

Answer 1

2cos^2x-cosx=2-secx
multiply both sides by cosx;
2cos^3x-cos^2x=2cosx-1;
2cos^3x-cos^2x-2cosx+1=0
put cosx=y;
2y^3-y^2-2y+1=0
(y-1)(2y^2+y-1)=0
(y-1)(2y^2+y-1)=0
(y-1)(y+1)(2y-1)=0
y=1,-1,1/2
x=0,180,60 deg.

Answer 2

Since sec(x) = 1/cos(x), multiply both sides by cos(x):

2 cos³(x) - cos²(x) = 2 cos(x) - 1

Factor the left side:

cos²(x) (2 cos(x) - 1) = 2 cos(x) - 1

Don't divide by (2 cos(x) - 1); instead, subtract (2 cos(x) - 1) from both sides:

cos²(x) (2 cos (x) - 1) - (2 cos(x) - 1) = 0

Factor:

(cos²(x) - 1)(2 cos(x) - 1) = 0

This tells us that either cos²(x) - 1 = 0 or 2 cos(x) - 1 = 0

Now find the solutions for x.

Answer 3

Everything on one side:

2cos^2 x - cos x - 2 + sec x = 0

Multiply everything by cos x (recalling sec x = 1/cos x):

2cos^3 x - cos^2 x - 2cos x + 1 = 0.

Now, set y = cos x:

2y^3 - y^2 - 2y + 1 = 0.

Solve for y. For each solution, replace y and solve for x in y = cos x. Then, check your answers (important!). Good luck.