sin^2x-cos^2=sinx

Question

 sin^2x-cos^2=sinx

 

please answer

Answer 1

sin^2x-cos^2x=sinx

sin^2x+cos^2x=1 is a trig identity

from this we get sin^2x-1=-cos^2x

substitute

sin^2x+sin^2x-1=sinx

2sin^2x-1=sinx

2sin^2x-sinx-1=0

factor

(2sinx+1)(sinx-1)=0

2sinx+1=0

2sinx=-1

sinx=-1/2

the sine is negative in the 3rd and 4th quadrants

looking on the unit circle:

x=7pi/6=210º and

x=11pi/6=330º

sinx-1=0

sinx=1

x=pi/2=90º

 

Answer 2

sin²x - cos²x = sinx

 

sin²x - (1 - sin²x) = sinx

 

2sin²x - sinx - 1 = 0

 

(2sinx + 1)(sinx - 1) = 0

 

sinx = -1/2 or sinx = 1

 

If sinx = -1/2, then x = 7π/6 + 2kπ or 11π/6 + 2kπ, k = 0, ±1, ±2,...

If sinx = 1, then x = π/2 + 2kπ

 

If you just need the solutions between 0 and 2π, they are

 

       π/2, 7π/6, and 11π/6