4cos(pi-x)(sin^2)x+cosx =0

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To solve the equation 4cos(pi-x)(sin^2)x + cosx = 0, we first need to simplify the expression.

Recall that cos(pi - x) = -cosx, so we can rewrite the equation as:

4(-cosx)(sin^2)x + cosx = 0

Now, let's expand the expression:

-4cosx(sin^2)x + cosx = 0

Using the trigonometric identity sin^2(x) = 1 - cos^2(x), we can rewrite the expression as:

-4cosx(1 - cos^2(x)) + cosx = 0

Expanding further:

-4cosx + 4cos^3(x) + cosx = 0

Combining like terms:

4cos^3(x) - 3cosx = 0

Now, let's factor out a common factor of cosx:

cosx(4cos^2(x) - 3) = 0

Setting each factor equal to zero:

cosx = 0

This gives us the solution x = pi/2 + n*pi for n being an integer.

4cos^2(x) - 3 = 0

Solving for cos(x):

cos^2(x) = 3/4

cos(x) = ±√(3)/2

This gives us the solutions x = π/6 + npi, 5π/6 + npi, where n is an integer.

Therefore, the solutions to the equation 4cos(pi-x)(sin^2)x + cosx = 0 are:

x = pi/2 + npi, π/6 + npi, 5π/6 + n*pi, where n is an integer.