Question 148293: Hi,
Question is solve the equation on the interval [0,2Pi0.
2cos^2x+sinx-2=0
Since cos^2x= 1-sin^2x I substituded this:
2(1-sin^2x) + sinx-2=0 to get
-2sin^2x + sinx=0 since -2 and +2 cancel eachother out. I am lost after this and the calculator hmm don't know whats worse figuring out the calculator or the math problem.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Question is solve the equation on the interval [0,2Pi].
2cos^2x+sinx-2=0
Since cos^2x= 1-sin^2x I substituded this:
2(1-sin^2x) + sinx-2=0 to get
-2sin^2x + sinx=0 since -2 and +2 cancel each other out.
2sin^2x - sinx = 0
Now factor to get:
sinx(2sinx-1) = 0
sinx = 0 or 2sinx = 1
sinx = 0 or sinx = 1/2
x = 0,pi,2pi or x = pi/6,(5/6)pi
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Cheers,
Stan H.
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