Asked by nemo
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
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
Answers
Answered by
oobleck
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.
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.
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