Find solution of the equation within the interval [0,2pi): 4-4sint=4(3^(1/2))cost

4 - 4sint = 4√3 cost

divide by 4

1 - sin(t) = √3 cos t

squaring on both sides

1 + sin^2(t) - 2sin t = 3 cos^2(t)

=> 1 + sin^2(t) - 2sin t = 3( 1 - sin^2(t))

=> 1 + sin^2(t) - 2sin t = 3 - 3sin^2(t))

=> 4sin^2(t) - 2sin t - 2 = 0

=> 2sin^2(t) - sin t - 1 = 0

=> ( 2sin t + 1)(sin t - 1) = 0

sin t = -1/2 and 1

t = π/2, 7π/6, 11π/6

Since we have squared in 3 rd step, we have to check for extraneous results.

7π/6 is extraneous, since LHS equals 6 and RHS equals - 6

Thus t = π/2 and 11π/6

4 - 4 * sin(t) = 4 * sqrt(3) * cos(t)
1 - sin(t) = sqrt(3) * cos(t)
1 - 2sin(t) + sin(t)^2 = 3 * cos(t)^2
1 - 2sin(t) + sin(t)^2 = 3 - 3sin(t)^2
4sin(t)^2 - 2sin(t) - 2 = 0
2sin(t)^2 - sin(t) - 1 = 0
sin(t) = (1 +/- sqrt(1 + 4 * 2 * 1)) / (2 * 2)
sin(t) = (1 +/- sqrt(9)) / 4
sin(t) = (1 +/- 3) / 4
sin(t) = -2/4 , 4/4
sin(t) = -1/2 , 1
t = 7pi/6 , 11pi/6 , pi/2

Test:

4 - 4 * sin(7pi/6) =>
4 - 4 * (-1/2) =>
4 + 2 =>
6

4 * sqrt(3) * cos(7pi/6) =>
4 * sqrt(3) * (-sqrt(3) / 2) =>
2 * -3 =>
-6

7pi/6 is not an answer


4 - 4 * sin(11pi/6) = 4 * sqrt(3) * cos(11pi/6)
4 - 4 * (-1/2) = 4 * sqrt(3) * sqrt(3) / 2
4 + 2 = 4 * 3 / 2
6 = 6

11pi/6 is an answer

4 - 4 * sin(pi/2) = 4 * sqrt(3) * cos(pi/2)
4 - 4 * 1 = 4 * sqrt(3) * 0
4 - 4 = 0
0 = 0

pi/2 is an answer


t = pi/2 , 11pi/6