find the solutions of the equation that are in the interval [0,2pi)

(a) sin2t+sint=0

(b)cost-sin2t=0

change the double angles to single angles.

sin2t+sint=0
2sint*cost+sint=0
sint(2cost+1)=0
and solve.

To find the solutions of the equation in the given interval [0, 2π), we can use the trigonometric identities and solve for the variable within the given range. Let's solve each equation step by step:

(a) sin(2t) + sin(t) = 0:

1. Use the double-angle identity for sine: sin(2t) = 2sin(t)cos(t).
Substitute the identity into the equation.
2sin(t)cos(t) + sin(t) = 0

2. Factor out sin(t) from both terms:
sin(t)(2cos(t) + 1) = 0

3. Set each factor equal to zero and solve for t:

i. sin(t) = 0
This occurs when t = 0, π, or 2π in the interval [0, 2π).

ii. 2cos(t) + 1 = 0
Subtract 1 from both sides:
2cos(t) = -1
Divide both sides by 2:
cos(t) = -1/2

To find the solutions for cos(t) = -1/2 within the given interval [0, 2π):
- The unit circle shows that cos(t) = -1/2 at 2 points: 2π/3 and 4π/3.
- However, 2π/3 is outside the range [0, 2π), so we discard it.
- Therefore, the only solution for cos(t) = -1/2 in the given interval is t = 4π/3.

Hence, the solutions to the equation sin(2t) + sin(t) = 0 in the interval [0, 2π) are t = 0, π, and 4π/3.

(b) cos(t) - sin(2t) = 0:

1. Use the double-angle identity for sine: sin(2t) = 2sin(t)cos(t).
Substitute the identity into the equation.
cos(t) - 2sin(t)cos(t) = 0
Factor out a common term of cos(t):
cos(t)(1 - 2sin(t)) = 0

2. Set each factor equal to zero and solve for t:

i. cos(t) = 0
This occurs when t = π/2 or 3π/2 in the interval [0, 2π).

ii. 1 - 2sin(t) = 0
Add 2sin(t) to both sides:
2sin(t) = 1
Divide both sides by 2:
sin(t) = 1/2

To find the solutions for sin(t) = 1/2 within the given interval [0, 2π):
- The unit circle shows that sin(t) = 1/2 at two points: π/6 and 5π/6.
- Both π/6 and 5π/6 are within the range [0, 2π).

Hence, the solutions to the equation cos(t) - sin(2t) = 0 in the interval [0, 2π) are t = π/2, 3π/2, π/6, and 5π/6.