SOLUTION: {{{(2cos^2x-3sinx=0}}} the cos is powered

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Question 684843: %282cos%5E2x-3sinx=0
the cos is powered
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
2cos%5E2%28x%29+-+3sin%28x%29=0
cos^2 = 1 - sin^2
2%281+-+sin%5E2%28x%29%29+-+3sin%28x%29=0
2sin%5E2%28x%29+%2B+3sin%28x%29+-+2+=+0
----------------
sub y for sin(x)
2y%5E+%2B+3y+-+2+=+0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B3x%2B-2+=+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=%283%29%5E2-4%2A2%2A-2=25.

Discriminant d=25 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-3%2B-sqrt%28+25+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%283%29%2Bsqrt%28+25+%29%29%2F2%5C2+=+0.5
x%5B2%5D+=+%28-%283%29-sqrt%28+25+%29%29%2F2%5C2+=+-2

Quadratic expression 2x%5E2%2B3x%2B-2 can be factored:
2x%5E2%2B3x%2B-2+=+%28x-0.5%29%2A%28x--2%29
Again, the answer is: 0.5, -2. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B3%2Ax%2B-2+%29

sin(x) = -2, ignore, no real numbers.
sin(x) = 0.5
x = pi/6 + 2n*pi, n = 0,1,2,3
x = 5pi/6 + 2n*pi, n = 0,1,2,3