1+sin X = 2cos^2 X

Question

how do I solve this?

Answer 1

1 + sin X = 2cos² X

 

1 + sin x = 2 (1 - sin²x)

1 + sin x = 2 - 2 sin²x

2 sin²x + sin x - 1 = 0

 

Let u = sin x

 

2u² + u -  1 = 0

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

u = -1 and 1/2

 

u = -1

sin x = -1

x = 3pi/2 ± 2npi, n=1, 2, 3, ...

 

u = 1/2

sin x = 1/2

x = pi/6 ± 2npi, and 5pi/6 ± 2npi,  n=1, 2, 3, ...

Answer 2

You want to get the equation in terms of sin(x).

 

1 + sinx = 2(1 - sin²x)

 

1 + sinx = 2 - 2sin²x

 

 

Move all terms to the left side of equation.

 

sin²x + sinx - 1 = 0

 

 

Now you have a quadratic equation.  Use the quadratic formula to solve for sin(x):

 

sin(x) = (-1 ± √(1 + 4)) / 2

 

sin(x) = (-1 ± √5) / 2

 

x = sin^-1[(-1 - √5) / 2]          and         x = sin^-1[(-1 + √5) / 2]