cot^2(3x)sin(2x)-sin(2x)=0

2sin^2(5x)+sin(5x)-1=0

For the first, take out a common factor of sin(2x) to get

sin(2x)[cot^2(3x) - 1] = 0
now you have a difference of squares for
sin(2x)(cot(3x - 1)(cot(3x) + 1) = 0
Can you take it from there?

I will do the second one:
Your equation is a quadratic which factors.
(2sin(5x) - 1)(sin(5x) + 1) = 0
sin(5x) = 1/2 or sin(5x) = -1

5x = 30º or 150º or 5x = 270º
x = 6º , 30º , 54º

I don't know what your domain was, but the period of sin(5x) is 360/5º or 72º
so by adding or subtracting multiples of 72 to each of my answers will yield as many answers as you want.

general solutions in radians :
pi/30 + k(2pi/5)
pi/6 + k(2pi/5)
3pi/10 + k(2pi/5) , where k is an integer.

Thank you! Would you mind explaining to me the k(2pi/5) thing? How do you know when to use that and how do you know when to use just k(pi/5) without the 2? Is the period always going to be the number in front of x (like in this case it was 5 because of 5x)?

To find the solutions to the equations cot^2(3x)sin(2x) - sin(2x) = 0 and 2sin^2(5x) + sin(5x) - 1 = 0, we can solve them step by step.

1. cot^2(3x)sin(2x) - sin(2x) = 0:
a. Factor out sin(2x): sin(2x)(cot^2(3x) - 1) = 0.
b. Set each factor equal to zero:
i. sin(2x) = 0.
To solve this, we need to find the values of x for which sin(2x) = 0. This happens when 2x is an integer multiple of π (0, ±π, ±2π, ±3π, ...).
Thus, 2x = nπ, where n is an integer.
Divide both sides by 2 to solve for x: x = nπ/2.
ii. cot^2(3x) - 1 = 0.
Rearrange the equation: cot^2(3x) = 1.
Take the square root of both sides: |cot(3x)| = 1.
There are two cases to consider, where cot(3x) is 1 or -1.
Case 1: cot(3x) = 1.
Solve: 3x = π/4 + nπ, where n is an integer.
Divide both sides by 3: x = (π/4 + nπ)/3.
Case 2: cot(3x) = -1.
Solve: 3x = 3π/4 + nπ, where n is an integer.
Divide both sides by 3: x = (π/4 + nπ)/3.

Therefore, the solutions to cot^2(3x)sin(2x) - sin(2x) = 0 are:
x = nπ/2, (π/4 + nπ)/3, and (π/4 + nπ)/3, where n is an integer.

2. 2sin^2(5x) + sin(5x) - 1 = 0:
This equation is a quadratic equation in terms of sin(5x). We can solve it using the quadratic formula.
a. Identify the coefficients A, B, and C in the general quadratic equation form Ax^2 + Bx + C = 0:
A = 2, B = 1, and C = -1.
b. Apply the quadratic formula: x = (-B ± √(B^2 - 4AC)) / (2A).
Plugging in the values: sin(5x) = [-(1) ± √((1)^2 - 4(2)(-1))] / (2(2)).
c. Simplify: sin(5x) = [-1 ± √(1 + 8)] / 4.
sin(5x) = [-1 ± √(9)] / 4.
sin(5x) = [-1 ± 3] / 4.
d. Two cases need to be considered.
Case 1: sin(5x) = (-1 + 3) / 4 = 2 / 4 = 1 / 2.
From the unit circle or trigonometric ratios, we can find one possible solution as arcsin(1/2) = π/6 + 2nπ.
Simplifying, we get 5x = π/6 + 2nπ, where n is an integer.
Divide both sides by 5: x = (π/6 + 2nπ)/5.
Case 2: sin(5x) = (-1 - 3) / 4 = -4 / 4 = -1.
From the unit circle or trigonometric ratios, we can find another possible solution as arcsin(-1) = -π/2 + 2nπ.
Simplifying, we get 5x = -π/2 + 2nπ, where n is an integer.
Divide both sides by 5: x = (-π/2 + 2nπ)/5.

Therefore, the solutions to 2sin^2(5x) + sin(5x) - 1 = 0 are:
x = (π/6 + 2nπ)/5 and (-π/2 + 2nπ)/5, where n is an integer.