If 2f(Sinx) + f(Cosx) =3Cos^2(x) find f ’(x) ?

Answer 1

Let x = -y
Then 2f(sin(-y)) + f(cos(-y)) = 3cos²(-y)
So
2f(-sin(y)) + f(cos(y)) = 3cos²(y)
or
2f(-sin(x)) + f(cos(x)) = 3cos²(x)
Subtracting this from the original, we see that
2f(sin(x)) = 2f(-sin(x))
or
f(x) = f(-x)

Now Let x = π/2+y.
Then 2f(sin(π/2+y)) + f(cos(π/2+y)) = 3cos²(π/2+y)
So
2f(cos(y)) + f(-sin(y)) = 3sin²(y)
or
2f(cos(x)) + f(-sin(x)) = 3sin²(x) which from the above is
2f(cos(x)) + f(sin(x)) = 3sin²(x)
Now adding this to the original equation, we get:
3f(sin(x)) + 3f(cos(x)) = 1

Therefore, using the chain rule we have:
3f'(sin(x))cos(x) + 3f'(cos(x))(-sin(x)) = 0 or
f'(sin(x))cos(x) - f'(cos(x)sin(x) = 0

Now let's go back to the original equation and differentiate:
2f'(sin(x))cos(x) - f'(cos(x))sin(x) = -6cos(x)sin(x)
or
f'(sin(x))cos(x) = -6cos(x)sin(x) or
f'(sin(x)) = -6sin(x)

That is to say,
f'(x) = -6x


Oh, and if you plug b-3x² back into the original equation, you can find that
f(x) = 2 - 3x²