Trig help... equation on the interval [0,2pi]?
2sin^2(x)=sin(x)
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To solve the equation 2sin^2(x) = sin(x) on the interval [0, 2π], we can rearrange the equation and solve for sin(x) first.
Step 1: Rearrange the equation:
2sin^2(x) - sin(x) = 0
Step 2: Factor out sin(x):
sin(x)(2sin(x) - 1) = 0
Now, set each factor equal to zero and solve for sin(x):
Factor 1: sin(x) = 0
To find the values of x where sin(x) = 0 on the interval [0, 2π], we can recall that sin(x) = 0 at x = 0 and x = π. Therefore, the solutions for sin(x) = 0 on the interval [0, 2π] are x = 0 and x = π.
Factor 2: 2sin(x) - 1 = 0
Solve for sin(x):
2sin(x) = 1
sin(x) = 1/2
To find the values of x where sin(x) = 1/2 on the interval [0, 2π], we can recall that sin(x) = 1/2 at x = π/6 and x = 5π/6. Therefore, the solutions for sin(x) = 1/2 on the interval [0, 2π] are x = π/6 and x = 5π/6.
Thus, the solutions for the equation 2sin^2(x) = sin(x) on the interval [0, 2π] are x = 0, x = π/6, x = π, and x = 5π/6.