To solve the equation sin(2x) sin(x) = 1 for 0° ≤ x ≤ 360°:
We use the double angle identity for sine: sin(2x) = 2sin(x)cos(x).
So, the equation becomes: 2sin(x)cos(x)sin(x) = 1.
Simplify and solve for sin(x):
2sin^2(x)cos(x) = 1.
Using the identity sin^2(x) = 1 - cos^2(x), we rewrite the equation:
2(1 - cos^2(x))cos(x) = 1.
Expanding and rearranging terms:
2cos(x) - 2cos^3(x) = 1.
2cos(x) = 1 + 2cos^3(x).
cos(x) = (1 + 2cos^3(x)) / 2.
Using the fact that cos^2(x) = 1 - sin^2(x), rewrite the equation in terms of sin(x):
cos(x) = (1 + 2(1 - sin^2(x))) / 2.
cos(x) = (1 + 2 - 2sin^2(x)) / 2.
cos(x) = (3 - 2sin^2(x)) / 2.
cos(x) = 3/2 - sin^2(x).
cos(x) + sin^2(x) = 3/2.
Using the Pythagorean identity sin^2(x) + cos^2(x) = 1, rewrite the equation as:
1 + cos(x) = 3/2.
cos(x) = 3/2 - 1.
cos(x) = 1/2.
Now, solve for x:
x = arccos(1/2).
x = 60°, 300°.
So, the solutions for 0° ≤ x ≤ 360° are x = 60° and x = 300°.
