Question

Solve 6cos^2x-sinx-4=0

Answer 1

6[cos²(x)] − sinx − 4 = 0
substitute [cos²(x)] = 1 − sin²(x)
substitute u = sin(x)
solve quadratic for "u"
... and so on

Answer 2

After solving for u, make sure that it lies between -1 and 1, because any other value of u cannot be equal to sin(x) or cos(x). Discard any solution that isn't in the interval.

Answer 3

But I thought cos2x=1-2sin^2x....is cox^2x=1-sin^2x as well?

Answer 4

\[ \sin^2 x + \cos^2 x = 1 \]
Rearrange to get cos²(x) in terms of sin²(x).

Answer 5

oh right okay thank you:)