How do you solve cos2x+sin2x=0

Question

This is a trig question solving for x using radians.

Answer 1

Question: solve cos2x+sin2x=0 .

cos2x+sin2x=0    --- rearrange

cos2x = -sin2x    --- divide both sides by cos2x

1 = -sin2x/tan2x, i.e.

tan2x = -1

Now tan(α) is positive if α is in the 1st quadrant or the 3rd quadrant.

But, since we have tan2x as being negative then 2x must lie in the 2nd quadrant or the 4th quadrant.

Since tan(-α) = -tan(α) and tan(π/4) = 1, then tan(-π/4) = -tan(π/4) = -1

i.e. 2x = -π/4, when 2x is in the 4th qaudrant

and 2x = 3π/4, when 2x is in the 3rd quadrant

Answer: x = -π/8, x = 3π/8