Question

sin^2x - cos^2x = -1/2

Answer 1

sin^2x - cos^2x = -1/2    /*2

 

2sin^2x - 2cos^2x = -1

 

2sin^2x - 2cos^2x = -sin^2x-cos^2x

 

2sin^2x - 2cos^2x + sin^2x + cos^2x = 0

 

3sin^2x - cos^2x = 0   /:cos^2x≠0

 

3tg^2x - 1 = 0

 

tg^2x = 1/3

 

tgx = ± √3/3

 

x= ± pi/6 + pik, k∈Z

 

Answer 2

$sin^2x - cos^2x = -frac{1}{2}$

$-(cos^2x - sin^2x) = -frac{1}{2}$

$cos2x =frac{1}{2}$

$2x=pmfrac{pi}{3}+2pi n$

$x=pmfrac{pi}{6}+pi n, nin Z$