solve for x: 4sin^2x= 4cosx+1
To solve the equation 4sin^2x = 4cosx + 1 for x, you can follow these steps:
Step 1: Use the identity sin^2x + cos^2x = 1
Rearranging the identity, you get sin^2x = 1 - cos^2x
Step 2: Substitute the identity from Step 1 into the original equation.
4(1 - cos^2x) = 4cosx + 1
Step 3: Distribute and simplify.
4 - 4cos^2x = 4cosx + 1
Step 4: Rearrange the equation to form a quadratic equation.
4cos^2x + 4cosx + 1 - 4 = 0
4cos^2x + 4cosx - 3 = 0
Step 5: Solve the quadratic equation.
You can solve it by factoring, completing the square, or using the quadratic formula. In this case, factoring may not be straightforward, so we will use the quadratic formula:
cosx = (-4 ± √(4^2 - 4(4)(-3))) / (2(4))
cosx = (-4 ± √(16 + 48)) / 8
cosx = (-4 ± √64) / 8
cosx = (-4 ± 8) / 8
Step 6: Simplify the solutions.
cosx = (-12/8) or cosx = (4/8)
cosx = -3/2 or cosx = 1/2
Step 7: Check for extraneous solutions.
Since the cosine function is only defined between -1 and 1, we can see that cosx = -3/2 is not a valid solution. However, cosx = 1/2 is valid.
Step 8: Solve for x.
Use the inverse cosine function (also known as arccos) to solve for x:
x = arccos(1/2)
Step 9: Find the principal value of x.
The principal range of the inverse cosine function is between 0 and π radians (or between 0 and 180 degrees). Therefore, x = arccos(1/2) gives us x = π/3 or x = 60°.
So, the solutions to the equation 4sin^2x = 4cosx + 1 are x = π/3 and x = 60°.