solve for x: 4sin^2x= 4cosx+1
To solve the equation 4sin^2x = 4cosx + 1, we can use the trigonometric identities to rewrite it in terms of a single trigonometric function.
First, let's recall the Pythagorean identity: sin^2x + cos^2x = 1.
Dividing the entire equation by 4, we get sin^2x = cosx/4 + 1/4.
Now, let's rewrite the right side of the equation in terms of sinx to eliminate the cosine term. Using the Pythagorean identity, we have cosx = √(1 - sin^2x).
Substituting this back into the equation, we get:
sin^2x = (√(1 - sin^2x))/4 + 1/4.
To simplify this, let's multiply the entire equation by 4:
4sin^2x = √(1 - sin^2x) + 1.
Square both sides of the equation to eliminate the square root:
(4sin^2x)^2 = (√(1 - sin^2x) + 1)^2.
Simplifying further:
16sin^4x = (1 - sin^2x) + 2√(1 - sin^2x) + 1.
Now, let's gather all the terms on one side of the equation:
16sin^4x + sin^2x - 2√(1 - sin^2x) = 0.
Now, we have a quadratic equation in terms of sinx. Let's substitute u = sin^2x to simplify the equation further:
16u^2 + u - 2√(1 - u) = 0.
This equation can be solved by factoring or using the quadratic formula. However, since it is a bit more complex, we can utilize numerical methods or a graphing calculator to solve for the value(s) of u.
Once we find the value(s) of u, we can substitute them back into u = sin^2x to find the corresponding values of x.