Question

Sin3x+2cos2x=2 how many values of x satisfy in the interval [-pi,7pi/2]

Answer 1

Wait just time for answer just time

Answer 2

sin3x - sinx = 1/2

2cos(2x)sin(x) = 1/2

cos(2x)sin(x) = 1/4

(1 - 2sin²(x))(sin(x)) = 1/4

sin(x) - 2sin³(x) - 1/4 = 0

2sin³(x) - sin(x) + 1/4 = 0

2u³ - u + ¼ = 0

This factors to:

¼ (2u - 1) (4u² + 2u - 1) = 0

Therefore u = ½, u = (-1 ± √5) / 4

sin(x) = 1/2

x = π/6 ± 2πk, x = 5π/6 ± 2πk

where k is any integer

sin(x) = (-1 ± √5) / 4