Answer:
π/14, 3π/14, 5π/14
Explanation:
sin(4θ) cos(2θ) = sin(5θ) cos(3θ)
Multiply both sides by 2:
2 sin(4θ) cos(2θ) = 2 sin(5θ) cos(3θ)
Use product to sum formula:
sin(4θ+2θ) + sin(4θ−2θ) = sin(5θ+3θ) + sin(5θ−3θ)
sin(6θ) + sin(2θ) = sin(8θ) + sin(2θ)
Simplify:
sin(6θ) = sin(8θ)
sin(8θ) − sin(6θ) = 0
Use sum to product formula:
2 sin(½(8θ−6θ)) cos(½(8θ+6θ)) = 0
2 sin(θ) cos(7θ) = 0
Solve:
sin(θ) = 0 or cos(7θ) = 0
θ = kπ
or
7θ = π/2 + kπ
θ = π/14 + k/7 π
θ = (2k+1)/14 π
0 < θ < π/2, so:
θ = π/14, 3π/14, 5π/14