SOLUTION: 4 cos^2 (4θ) + 4 sin (4θ) = 3

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Question 1029080: 4 cos^2 (4θ) + 4 sin (4θ) = 3
Found 2 solutions by Fombitz, Alan3354:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Substitute u=cos%284%2Atheta%29
4u%5E2%2B4u-3=0
%282u-1%29%282u%2B3%29=0
Two "u" solutions:
2u-1=0
2u=1
u=1%2F2
So then,
cos%284%2Atheta%29=1%2F2
4%2Atheta=60
4%2Atheta=60%2B360=420
4%2Atheta=60%2B720=780
4%2Atheta=60%2B1080=1140
So,
theta=15
theta=105
theta=195
theta=285
and
4%2Atheta=300
4%2Atheta=300%2B360=660
4%2Atheta=300%2B720=1020
4%2Atheta=300%2B1080=1380
So,
theta=75
theta=165
theta=255
theta=345
8 solutions
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.
.
2u%2B3=0
2u=-3
u=-3%2F2
cos%284%2Atheta%29=-3%2F2
Since cosine can never equal this value, this u solution does not yield a theta solution.

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
4cos^2(t) + 4sin(t) = 3
4(1 - sin^2(t)) + 4sin(t) = 3
4sin^2 - 4sin - 1 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 4x%5E2%2B-4x%2B-1+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-4%29%5E2-4%2A4%2A-1=32.

Discriminant d=32 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--4%2B-sqrt%28+32+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-4%29%2Bsqrt%28+32+%29%29%2F2%5C4+=+1.20710678118655
x%5B2%5D+=+%28-%28-4%29-sqrt%28+32+%29%29%2F2%5C4+=+-0.207106781186548

Quadratic expression 4x%5E2%2B-4x%2B-1 can be factored:
4x%5E2%2B-4x%2B-1+=+%28x-1.20710678118655%29%2A%28x--0.207106781186548%29
Again, the answer is: 1.20710678118655, -0.207106781186548. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+4%2Ax%5E2%2B-4%2Ax%2B-1+%29

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sin(t) =~ -0.2071
t =~ 192, 348 degs