1) sin(π + 3x) = 0 Since sin(π + θ) = sin(θ), we can rewrite the equation as: sin(3x) = 0
To find the values of x, we need to determine when sin(3x) equals zero. Sin(x) = 0 when x is a multiple of π, i.e., x = 0, π, 2π, 3π, ...
So, 3x = nπ, where n is an integer. Thus, x = nπ/3, where n is an integer.
2) cos^2(x) = sin^2(x) Using the Pythagorean identity sin^2(x) + cos^2(x) = 1, we can rewrite the equation as: 1 - sin^2(x) = sin^2(x) 1 = 2sin^2(x) sin^2(x) = 1/2
Taking the square root of both sides, we get: sin(x) = ±√(1/2) sin(x) = ±1/√2 Since sin(π/4) = 1/√2 and sin(3π/4) = -1/√2, the solutions are: x = π/4 + 2πn, x = 3π/4 + 2πn (where n is an integer)
1) sin(π + 3x) = 0
Since sin(π + θ) = sin(θ), we can rewrite the equation as:
sin(3x) = 0
To find the values of x, we need to determine when sin(3x) equals zero. Sin(x) = 0 when x is a multiple of π, i.e., x = 0, π, 2π, 3π, ...
So, 3x = nπ, where n is an integer.
Thus, x = nπ/3, where n is an integer.
2) cos^2(x) = sin^2(x)
Using the Pythagorean identity sin^2(x) + cos^2(x) = 1, we can rewrite the equation as:
1 - sin^2(x) = sin^2(x)
1 = 2sin^2(x)
sin^2(x) = 1/2
Taking the square root of both sides, we get:
sin(x) = ±√(1/2)
sin(x) = ±1/√2
Since sin(π/4) = 1/√2 and sin(3π/4) = -1/√2, the solutions are:
x = π/4 + 2πn, x = 3π/4 + 2πn (where n is an integer)