SOLUTION: Find all the angles between 0� and 360� inclusive which satisfy the equation 4sin^2(x) - 7sinxcosx

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Question 1056152: Find all the angles between 0� and 360� inclusive which satisfy the equation 4sin^2(x) - 7sinxcosx - 5cos^2(x) + 2 = 0
Answer by htmentor(1343) (Show Source):
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The key in solving this problem is to recognize that the number 2 can be expressed as 2sin^2(x) + 2cos^2(x)
Then the expression becomes 6sin^2(x) - 7sin(x)cos(x) - 3cos^2(x) = 0
Dividing through by cos^2(x) gives us a quadratic in tan(x):
6tan^2(x) - 7tan(x) - 3 = 0
For simplicity, let z = tan(x)
6z^2 - 7z - 3 = 0
The LHS can be factored as (3z+1)(2z-3)
So we need to solve 3z+1 = 0 and 2z-3 = 0
3z+1 = 0 -> z = -1/3
x = arctan(-1/3) = -0.3218; tangent function has periodicity of pi, so the solutions between 0 and 2*pi are -0.3218+pi = 2.82 and -0.3218+2*pi = 5.96
2z-3 = 0 -> z = 3/2
x = arctan(3/2)
Solutions are x = 0.983 and 4.124
Graph is below:

SOLUTION: Find all the angles between 0� and 360� inclusive which satisfy the equation 4sin^2(x) - 7sinxcosx - 5cos^2(x) + 2 = 0