SOLUTION: Find all values of x between 0 and 360 degrees that satisfy the equation 6cos(x)^2+5sin(x)=2. Give your answer to the nearest tenth of a degree.
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Question 423510: Find all values of x between 0 and 360 degrees that satisfy the equation 6cos(x)^2+5sin(x)=2. Give your answer to the nearest tenth of a degree. Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! Find all values of x between 0 and 360 degrees that satisfy the equation 6cos(x)^2+5sin(x)=2. Give your answer to the nearest tenth of a degree.
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6cos^2x+5sinx=2
6(1-sin^2x)+5sinx-2=0
6-6sin^2x+5sinx-2=0
-6sin^2x+5sinx+4=0
6sin^2x-5sinx-4=0
solve by factoring this quadratic equation
(3sinx-4)(2sinx+1)=0
3sinx-4=0
3sinx=4
sinx=4/3 (reject,sinx<=1)
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2sinx+1=0
sinx=-1/2 (sin is negative in 3rd and 4th quadrants)
ans:
x=7pi/6 and 11pi/6
or 210 deg. and 330 deg.
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