The solutions for the equation 2 sin (9x) + 1 = 0 are all values of x = π/3 + 2πk, where k is any integer.
Isolate sin(9x):
Subtract 1 from both sides: 2 sin(9x) = -1.
Divide both sides by 2: sin(9x) = -1/2.
Find solutions for sin(9x) = -1/2:
Recall that sin(π/3) = √3/2 and sin(5π/3) = -√3/2.
Therefore, possible solutions for 9x are either π/3 or 5π/3.
Adjust solutions for periodicity of sine:
Since sine is periodic with period 2π, its values repeat every 2π units.
Add multiples of 2π to the solutions found in step 2 to obtain all possible solutions:
x = π/3 + 2πk, where k is any integer.
x = 5π/3 + 2πk, where k is any integer.
Combine solutions:
Both expressions represent the same set of solutions because:
(π/3 + 2πk) - (5π/3 + 2πk) = -4π/3, which is a multiple of 2π.
Therefore, the general solution is:
x = π/3 + 2πk, where k is any integer.
Question:-
Find all solutions of the equation
2 sin (9x) + 1 = 0