within the domain : (-p # x #p ), solve for x 2[(sinx)^4 + (cosx)^4) =1

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Originally Posted by white_cap within the domain : (-p # x #p ), solve for x 2[(sinx)^4 + (cosx)^4) =1 Your domain got screwed up a bit. Anyway: etc. Write this out as a polynomial in sin(x) and solve. -Dan

Hello, white_cap! Another approach . . . Solve for Divide by 2: . Add and subtract . . . . And we have: . . . Hence: . Therefore: .