Can somebody please help me figure these equations?

Question

1.) cos^2x=cosx

2.) 2cosx+radical3=0

3.) 4sin^2x=1

4.)2sin^2x+sinx=1

5.)cos^2x+2cosx=3

Answer 1

Hi, I'm going to assume that answers are ok in degrees (since it's just easier to type out), and that you're only looking for angles between 0° and 360°.

Also, many of the trig. equations listed have more than one solution.  The first can be found using the inverse trig. function on a calculator.  For example, if sinx = 1/2, then x = sin^-1 (1/2) = 30°.  The other solution(s) can be found using the following relationships:

sinx = sin(180° - x)

cosx = cos(360° - x)

I will be referring to them quite a bit.

1.)  cos^2x = cosx

cos^2x - cosx = 0  <-- rearrange

cosx(cosx - 1) = 0  <-- factor

thus, cosx = 0 or cosx = 1

if cosx = 0, then x = 90° or x = 360° - 90° = 270°

if cosx = 1, then x = 0° or x = 360° - 0° = 360°

therefore, x = 0°, 90°, 270°, or 360°

2.)  2cosx + √3 = 0

2cosx = -√3

cosx = -√3 ÷ 2

thus, x = 150° or x = 360° - 150° = 210°

therefore, x = 150° or 210°

3.)  4sin^2x = 1

sin^2x = 1/4

sinx = +/- 1/2  <-- take square root of both sides

if sinx = 1/2, then x = 30° or x = 180° - 30° = 150°

if sinx = -1/2, then x = -30° = 330°, or x = 180 - (-30°) = 210°

therefore, x = 30°, 150°, 210°, or 330°

4.)  2sin^2x + sinx = 1

2sin^2x + sinx - 1 = 0

(sinx + 1)(2sinx - 1) = 0  <-- factor

thus, sinx = -1 or sinx = 1/2

if sinx = -1, then x = -90° = 270°

if sinx = 1/2, then x = 30° or x = 180° - 30° = 150°

therefore, x = 30°, 150°, or 270°

5.)  cos^2x + 2cosx = 3

cos^2x + 2cosx - 3 = 0

(cosx + 3)(cosx - 1) = 0  <-- factor

thus, cosx = -3 or cosx = 1

if cosx = -3, then there is no solution

if cosx = 1, then x = 0° or x = 360° - 0° = 360°

therefore, x = 0° or 360°

I hope that helps.  Don't forget, you can always check the answer(s) to any equation by substituting the value(s) for the variable back into the original equation.