sin^2(x)=6(cos(x)+1)
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sin^2 x = 6(cosx + 1)
1 - cos^2 x = 6cosx + 6
0 = cos^2x + 6cosx + 5
0 = (cosx + 5)(cosx + 1)
cosx = - 5 does not exist
cos x = - 1
x = npi
1 - cos^2 x = 6cosx + 6
0 = cos^2x + 6cosx + 5
0 = (cosx + 5)(cosx + 1)
cosx = - 5 does not exist
cos x = - 1
x = npi
-
Use the Pythagorean Identity: cos² x + sin² x = 1, so sin² x = 1 - cos² x
1 - cos² x = 6(cos x) + 6 .... substituting on the left and multiplying out the right
cos² x + 6(cos x) + 5 = 0 ..... move all terms to the left
If you treat (cos x) as a variable, that's a quadratic. It factors as:
[(cos x) + 1] [(cos x) + 5] = 0
The zero product property says that this is true only when cos x = -1 or cos x = -5. But cos x is never -5, so only the first solution is possible for real values of x:
cos x = -1
x = π + 2πn .... for any integer n
1 - cos² x = 6(cos x) + 6 .... substituting on the left and multiplying out the right
cos² x + 6(cos x) + 5 = 0 ..... move all terms to the left
If you treat (cos x) as a variable, that's a quadratic. It factors as:
[(cos x) + 1] [(cos x) + 5] = 0
The zero product property says that this is true only when cos x = -1 or cos x = -5. But cos x is never -5, so only the first solution is possible for real values of x:
cos x = -1
x = π + 2πn .... for any integer n