SOLUTION: Can someone please help me solve these two problems for THETA θ sin^4 2θ - 2 sin^2 2θ = -1 (3tan 3θ)^2 - 27 = 0

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Question 1122857: Can someone please help me solve these two problems for THETA θ
sin^4 2θ - 2 sin^2 2θ = -1
(3tan 3θ)^2 - 27 = 0
Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
.
sin%5E4%282theta%29 - 2+sin%5E2%282theta%29 = -1  <====>  (is equivalent to) 


sin%5E4%282theta%29+-+2%2Asin%5E2%282theta%29+%2B+1 = 0  <====>  (is equivalent to)


%28sin%5E2%282theta%29+-1%29%5E2 = 0  <====>  (is equivalent to)


sin%5E2%282theta%29-1 = 0  <====>  (is equivalent to) 


sin%5E2%282theta%29 = 1   <====>  (is equivalent to)


sin%282theta%29 = +/- 1.


    Case 1.  sin%282theta%29 = 1   is equivalent to  2theta = pi%2F2, pi%2F2+%2B-+2pi, pi%2F2+%2B-+4pi, . . . , pi%2F2+%2B-+2n%2Api,  where n is any integer.

             General solution for this case is  theta = pi%2F4+%2B+n%2Api,  where n is any integer.



    Case 2.  sin%282theta%29 = -1   is equivalent to  2theta = %283pi%29%2F2, %283pi%29%2F2+%2B-+2pi, %285pi%29%2F2+%2B-+4pi, . . . , %283pi%29%2F2+%2B-+2n%2Api,  where n is any integer.

             General solution for this case is  theta = %283pi%29%2F4+%2B+n%2Api,  where n is any integer



Answer.  Any angle of the set  pi%2F4%2Bk%2Api, %283pi%29%2F4%2Bk%2Api  is the solution, where k is any integer.

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