The Student Room Group
Reply 1
Undry1
Solve for values of x, in degrees, in the range 0-360°,

5sin2x + 2cosx(2+2cos2x) = 0

im ending up with x = 90,270 but not sure. thanks


Your answers are correct.
Reply 2
Undry1
5sin2x + 2cosx(2+2cos2x) = 0


5sin2x + 2cosx(2+2cos2x) = 0
5sin2x + 4cosx + 4cos[x]cos[2x] = 0
5sin2x + 4cosx + 4cos[x]cos[2x] = 0

5(2sinxcosx) + 4cosx + 4cos[x](1-2sin²x) = 0
10sin[x]cos[x] + 4cosx + 4cosx - 8cos[x]sin²x = 0
10sin[x]cos[x] + 8cosx - 8cos[x]sin²x = 0
10sin[x] + 8 - 8sin²x = 0
8sin²x - 10sinx - 8 = 0

Let s = sinx

8s² - 10s - 8 = 0
solve this to get s ≈ 1.8042 and -0.554

sinx = -0.554
x = -33.64° etc.
Ricki
5sin2x + 2cosx(2+2cos2x) = 0
5sin2x + 4cosx + 4cos[x]cos[2x] = 0
5sin2x + 4cosx + 4cos[x]cos[2x] = 0

5(2sinxcosx) + 4cosx + 4cos[x](1-2sin²x) = 0
10sin[x]cos[x] + 4cosx + 4cosx - 8cos[x]sin²x = 0
10sin[x]cos[x] + 8cosx - 8cos[x]sin²x = 0
10sin[x] + 8 - 8sin²x = 0
8sin²x - 10sinx - 8 = 0

Let s = sinx

8s² - 10s - 8 = 0
solve this to get s ≈ 1.8042 and -0.554

sinx = -0.554
x = -33.64° etc.

Out of range. :frown:
Reply 4
there are two other solutions...

5sin2x + 2cosx(2+2cos2x) = 0
5(2sinxcosx) + 4cosx + 4cosx(1 + 4(2cos^2(x) - 1)) = 0
10sinxcosx + 2(cosx)^3 = 0
cosx(5sinx + 4(cosx)^2) = 0

...
Reply 5
Widowmaker
Out of range. :frown:


x = 180° + 33.64 = 213.64°
Dekota
x = 180° + 33.64 = 213.64°

:rolleyes:
I knew that, just teasing.