SOLUTION: {{{3*cos(beta) + 3}}} = {{{2*sin^2(beta)}}}

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Question 1030398: 3%2Acos%28beta%29+%2B+3 = 2%2Asin%5E2%28beta%29
Answer by ikleyn(52800) About Me  (Show Source):
You can put this solution on YOUR website!
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3%2Acos%28beta%29+%2B+3 = 2sin%5E2%28beta%29.

Replace  sin%5E2%28beta%29  by  1+-+cos%5E2%28beta%29.  You will get

3%2Acos%28beta%29+%2B+3 = 2%2A%281+-+cos%5E2%28beta%29%29,   or

3%2Acos%28beta%29+%2B+3 = 2+-+2%2Acos%5E2%28beta%29,   or

2%2Acos%5E2%28beta%29+%2B+3%2Acos%28beta%29+%2B+1 = 0,   or  (after factoring left side)

%282%2Acos%28beta%29+%2B+1%29%2A%28cos%28beta%29+%2B+1%29 = 0.


It gives you two equations.


1)  2%2Acos%28beta%29+%2B+1 = 0  --->  cos%28beta%29 = -1%2F2  --->  beta = 2pi%2F3+%2B+2k%2Api  or  beta = 4pi%2F3+%2B+2k%2Api,  k = 0, +/-1, +/-2, . . . 


2)  cos%28beta%29+%2B+1 = 0  --->  cos%28beta%29 = -1  --->  beta = pi+%2B+2k%2Api = %282k%2B1%29%2Api,  k = 0, +/-1, +/-2, . . .