Please I'm stuck in trigonometric equation

[Johnny Blaze]

Solve for x of:

4 Sin^2 x Cos x + 2 Sin^2x - 2 Cos x - 1 = 0

Here's my working

4(1-Cos^2 x) Cos x + 2 Sin^2 x - 2 Cos x - (Sin^2 x + Cos ^2 x) = 0

4 Cos x - 4 Cos^3 x + 2 Sin^2 x - 2 Cos x - Sin^2 x - Cos ^2 x = 0

4 Cos x - 4 Cos^3 x + 2 (1 - Cos^2 x) - 2 Cos x - (1 - Cos^2 x) - Cos ^2 x = 0

4 Cos x - 4 Cos^3 x + 2 - 2 Cos^2 x - 2 Cos x - 1 + Cos ^2 x - Cos^2 x = 0

4 Cos^3 x + 2 Cos^2 x - 2 Cos x - 1 = 0

Now I'm stuck...please help me. Thank you

 

[Deleted member 4993]

Solve for x of:

4 Sin^2 x Cos x + 2 Sin^2x - 2 Cos x - 1 = 0

Here's my working

4(1-Cos^2 x) Cos x + 2 Sin^2 x - 2 Cos x - (Sin^2 x + Cos ^2 x) = 0

4 Cos x - 4 Cos^3 x + 2 Sin^2 x - 2 Cos x - Sin^2 x - Cos ^2 x = 0

4 Cos x - 4 Cos^3 x + 2 (1 - Cos^2 x) - 2 Cos x - (1 - Cos^2 x) - Cos ^2 x = 0

4 Cos x - 4 Cos^3 x + 2 - 2 Cos^2 x - 2 Cos x - 1 + Cos ^2 x - Cos^2 x = 0

4 Cos^3 x + 2 Cos^2 x - 2 Cos x - 1 = 0

Now I'm stuck...please help me. Thank you

4 Cos^3 x + 2 Cos^2 x - 2 Cos x - 1 = 0

2*cos^2(x) * [2*cos(x)+1] -1 * [2*cos(x)+1] = 0

[2*cos^2(x) -1] * [2*cos(x)+1] = 0

Continue....

 

[Johnny Blaze]

Aww...I got it! Thank you so much!

 

[Steven G]

Aww...I got it! Thank you so much!

Great! I would suggest that since you are trying to only have powers osfcos(x) in your equation NOT to convert 1 into sin^2x + cos^2x as that introduces sin^2x, something that you do not want. Note that in the end that you ended up with +1 anyways. Your initial step should have been to convert both sin^2x to 1-cos^2x and continue as you did.