already posted these questions but i could not find the post
find all solutions from 0 degrees to 360 degrees
1. 4 sin^2x – 1 = 0
2. 2 sin^2x + sin x = 1
3. 2 sin^2x + 7 sin x = 4
4. cos x sin x + sin x = 0
5. 2 sin^2x - 1 = 0
c'mon -- these are basically algebra I, then a bit of trig knowledge.
1. 4 sin^2x – 1 = 0
(2sinx-1)(2sinx+1) = 0
so, sinx = ±1/2
2. 2 sin^2x + sin x = 1
2sin^2x + sinx - 1 = 0
(2sinx-1)(sinx+1) = 0
sinx = -1 or 1/2
3. 2 sin^2x + 7 sin x = 4
2sin^2x + 7sinx - 4 = 0
(2sinx-1)(sinx+4) = 0
sinx = -4 or 1/2
4. cos x sin x + sin x = 0
sinx(cosx+1) = 0
sinx = 0
or cosx = -1
5. 2 sin^2x - 1 = 0
(√2 sinx + 1)(√2 sinx - 1) = 0
so sinx = ±1/√2
Now just find all the angles in the domain where the sin/cos functions attain these values.
They are just the standard exact values you have learned on the unit circle.
To find all solutions for these trigonometric equations in the given range of 0 degrees to 360 degrees, we can follow these steps:
1. Solve the equation algebraically by manipulating it to isolate the trigonometric function. We need to transform the equation to a form like sin x = a or cos x = a.
2. Use the inverse trigonometric functions (sin⁻¹, cos⁻¹, tan⁻¹) to find the principal solution(s) within the range 0 to 360 degrees.
3. Determine whether there are any additional solutions using periodicity and symmetry of the trigonometric functions.
Let's go through each equation and apply these steps to find the solutions:
1. 4 sin^2x – 1 = 0:
- Begin by isolating sin^2x: 4 sin^2x = 1.
- Divide both sides by 4: sin^2x = 1/4.
- Take the square root of both sides: sin x = ±1/2.
- The principal solutions within the given range are sin⁻¹(1/2) = 30 degrees and sin⁻¹(-1/2) = 210 degrees.
- Since sine is positive in the first and second quadrants, we can add multiples of 360 degrees to these solutions.
- Therefore, the solutions are 30 degrees, 210 degrees, 390 degrees (30 + 360), and 570 degrees (210 + 360).
2. 2 sin^2x + sin x = 1:
- Rearrange the equation to get 2 sin^2x + sin x - 1 = 0.
- This equation can be factored: (2 sin x - 1)(sin x + 1) = 0.
- Set each factor equal to zero and solve:
- 2 sin x - 1 = 0 → sin x = 1/2 (principal solution: sin⁻¹(1/2) = 30 degrees)
- sin x + 1 = 0 → sin x = -1 (principal solution: sin⁻¹(-1) = 270 degrees)
- Applying multiples of 360 degrees to these solutions, we get 30 degrees, 210 degrees (30 + 180), 270 degrees, and 450 degrees (270 + 180).
3. 2 sin^2x + 7 sin x = 4:
- Rewrite the equation as 2 sin^2x + 7 sin x - 4 = 0.
- This quadratic equation can be factored: (2 sin x - 1)(sin x + 4) = 0.
- Set each factor equal to zero and solve:
- 2 sin x - 1 = 0 → sin x = 1/2 (principal solution: sin⁻¹(1/2) = 30 degrees)
- sin x + 4 = 0 → sin x = -4 (out of the range of -1 to 1, so there are no solutions from this factor)
- The solutions are 30 degrees and 210 degrees (30 + 180).
4. cos x sin x + sin x = 0:
- Factor out sin x from this equation: sin x (cos x + 1) = 0.
- Set each factor equal to zero and solve:
- sin x = 0 (multiple solutions: 0 degrees and 180 degrees)
- cos x + 1 = 0 → cos x = -1 (principal solution: cos⁻¹(-1) = 180 degrees)
- The solutions are 0 degrees, 180 degrees, 360 degrees (180 + 180), and 540 degrees (360 + 180).
5. 2 sin^2x - 1 = 0:
- Move the constant term to the other side: 2 sin^2x = 1.
- Divide both sides by 2: sin^2x = 1/2.
- Take the square root of both sides: sin x = ±√(1/2).
- The principal solutions within the given range are sin⁻¹(√(1/2)) = 45 degrees and sin⁻¹(-√(1/2)) = 315 degrees.
- The solutions are 45 degrees, 225 degrees (45 + 180), 315 degrees, and 495 degrees (315 + 180).
Those are the solutions for each equation within the range of 0 degrees to 360 degrees.