Please help!!!!!!!!!!! Find all solutions in the interval [0,2π).
7. 2 sin^2x=sin x
Please answer asap
To find all the solutions to the equation 2sin^2(x) = sin(x) in the interval [0, 2π), we can follow these steps:
1. Write the equation in standard form: 2sin^2(x) - sin(x) = 0.
2. Notice that this is a quadratic equation in terms of sin(x). Let's create an equation of the form Ax^2 + Bx + C = 0. In this case, A = 2, B = -1, and C = 0.
3. Factor out the common term, sin(x): sin(x)(2sin(x) - 1) = 0.
4. Set each factor equal to zero and solve for x:
a. sin(x) = 0:
Since we are looking for solutions in the interval [0, 2π), we know that sin(x) = 0 at x = 0, π, and 2π.
b. 2sin(x) - 1 = 0:
Solve for sin(x): sin(x) = 1/2.
Based on the unit circle, sin(x) = 1/2 at x = π/6 and x = 5π/6.
5. Consolidate all the solutions we found:
The solutions are x = 0, π/6, π, 5π/6, and 2π.
Note: When solving trigonometric equations, it is crucial to check the initial equation's original domain and restrictions. In this case, the equation was defined in the interval [0, 2π), so we only looked for solutions within that range.