Question: Find the values of x in the interval 0<x<360 for which Sin2x + Cos2x =0
I have attempted to square and then sub in using the identity: sin^2x + cos^2x = 1, but this just ends with 1 as both sin and cos just cancel each other out, and I don’t see a clear way of using tan x=sin x/ cos x .
Is there something that I am missing?
cos2x = -sin2x
-tan2x =1, 1+tan2x =0
then it may get the answer
Sin2x = 2sinxcosx
Cos2x = cosx^2 - sinx^2
Cos2x = 2cosx^2 - 1
Okay, I see what you did. According to the book I am working from the first x value is 67.5 degrees, therefore the solution above doesn't work out as it gives an x value of -22.5
In the case of the negative value of the tangent region, we can consider two quadrants (II and IV) such that quadrant II (sine region) is 90+(-22.5) = 67.5 and quadrant IV (cosine region) is 360+(-22.5) = 337.5.