SOLUTION: Solve the equation 3cosx + 4sinx = 2, for values of x between 0° and 360°

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Question 1152416: Solve the equation 3cosx + 4sinx = 2, for values of x between 0° and 360°
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

3cos%28x%29+%2B+4sin%28x%29+=+2, for values of x , 0+%3Cx%3C+360°

3cos%28x%29+%2B+4sin%28x%29+=+2

3cos%28x%29+=+2-4sin%28x%29 ......square both sides

%283cos%28x%29+%29%5E2=+%282-4sin%28x%29%29%5E2

9cos%5E2%28x%29+=+16sin%5E2%28x%29-16sin%28x%29%2B4

9cos%5E2%28x%29+%2B16sin%28x%29-16sin%5E2%28x%29-4=0.......use identity cos%5E2%28x%29=+1-sin%5E2%28x%29

9%281-sin%5E2%28x%29%29+%2B16sin%28x%29-16sin%5E2%28x%29-4=0

9-9sin%5E2%28x%29%29+%2B16sin%28x%29-16sin%5E2%28x%29-4=0

-25sin%5E2%28x%29%29+%2B16sin%28x%29%2B5=0.........let sin+%28x+%29=u

-25u%5E2+%2B16u%2B5=0............use quadratic formula

u=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a

u=%28-16%2B-sqrt%2816%5E2-4%28-25%29%2A5%29%29%2F%282%28-25%29%29

u=%28-16%2B-sqrt%28256%2B500%29%29%2F%28-50%29

u=%28-16%2B-sqrt%28756%29%29%2F%28-50%29

u=%28-16%2B-2sqrt%28189%29%29%2F%28-50%29

u=-%28-8%2B-sqrt%28189%29%29%2F25

u=-%28-8%2B-sqrt%289%2A21%29%29%2F25

u=-%28-8%2B-3sqrt%2821%29%29%2F25

use sin+%28x+%29=u

sin+%28x+%29=-%28-8%2B-3sqrt%2821%29%29%2F25

solutions:
sin+%28x+%29=8%2F25-3sqrt%2821%29%2F25
or
sin+%28x+%29=8%2F25%2B3sqrt%2821%29%29%2F25

then,
x=360-sin%5E-1%288%2F25-3sqrt%2821%29%2F25%29+
x=180-sin%5E-1%288%2F25%2B3sqrt%2821%29%2F25%29

solutions in decimal form:
x=-0.23198+%2B360=>highlight%28x=2pi-0.23198%29+
x=180-1.05501=>highlight%28x=pi-1.05501%29