Solve the following equation for all solutions: sin(11x)cos(6x)-cos(11x)sin(6x)=√3/2

Answer 1

sin ( A - B ) = sin A ∙ cos B - cos A ∙ sin B

that's why

sin ( 11 x ) ∙ cos ( 6 x ) - cos ( 11 x ) ∙ sin( 6 x ) = sin ( 11 x - 6 x ) = sin 5 x = √3 / 2

sin 5 x = √3 / 2

5 x = sin⁻¹ ( √3 / 2 ) = arcsin ( √3 / 2 )

The solutions of this equuation are:

5 x = π / 3 and 5 x = 2 π / 3

Period of sin x = 2 π , so the solutions of equuation sin 5 x = √3 / 2 are:

5 x = π / 3 + 2 π n and 5 x = 2 π / 3 + 2 π n

where:

n = ± 1 , ± 2 , ± 3...

Divide both sides by 5

x = π / 15 + 2 π n / 5 and x = 2 π / 15 + 2 π n / 5

Answer 2

sin(11x)cos(6x)-cos(11x)sin(6x)=√3/2
did you notice the pattern on the left?
sin(A-B) = sinAcosB - cosAsinB

so
sin(11x)cos(6x)-cos(11x)sin(6x)=√3/2
sin(11x - 6x) = √3/2
sin 5x = √3/2
5x = 60° or 120°
x = 15 or 24°
the period of sin 5x = 360/5° = 72°

you asked for all solutions, which is not possible, since we have an infinite number of solutions,
the general solution would be
x = 15° + 72k° or 24° + 72k°, where k is an integer, eg. 240° would be a solution

in case you want radians:
x = π/12 + (2π/5)k or 2π/15 ° + (2π/5)k °