THE VALUES OF X THAT ARE SOLUTIONS TO THE EQUATION [COSX]^2 = SIN2X IN THE INTERVAL 0 TO PI??
(cosx)^2 = 2sinxcox
(cosx)^2 - 2sinxcosx = 0
cosx(cosx - 2sinx) = 0
cosx = 0 or cosx = 2sinx
if cosx=0 then x = pi/2
if cosx = 2sinx
sinx/cosx = 1/2
tanx = 1/2
x = .46365
To find the values of x which are solutions to the equation [cos(x)]^2 = sin(2x) in the interval 0 to π, let's break it down step by step:
First, we'll simplify the equation using trigonometric identities:
1. Recall that sin(2θ) = 2sin(θ)cos(θ), so we can rewrite sin(2x) as 2sin(x)cos(x).
Therefore, the equation becomes: [cos(x)]^2 = 2sin(x)cos(x).
Next, let's manipulate the equation further:
2. Divide both sides of the equation by cos(x):
[cos(x)]^2 / cos(x) = 2sin(x)cos(x) / cos(x).
Simplifying, we get: cos(x) = 2sin(x).
Now, let's break this down into two separate equations:
3a. Equation 1: cos(x) = 0
Whenever cos(x) = 0, x must be equal to π/2 or 3π/2. However, we need to check if these values fall within the given interval of 0 to π. The value π/2 does fall within the interval, but 3π/2 is outside the interval, so we only consider x = π/2.
3b. Equation 2: 2sin(x) = cos(x)
We can rearrange this equation to: 2sin(x) - cos(x) = 0.
Applying the Pythagorean identity (sin^2(x) + cos^2(x) = 1), we get: 2sin(x) - √(1 - sin^2(x)) = 0.
Squaring both sides, we obtain: 4sin^2(x) - 4sin^2(x) + 1 = 0.
Simplifying further, we have: sin^2(x) + 1 = 0, which is not possible since sin^2(x) is always non-negative.
Therefore, there are no solutions to Equation 2 within the given interval.
In conclusion, the only solution within the interval 0 to π is x = π/2.