Find all solutions of the equation in the interval (0, 2π]. 1+sin x = 2 cos^2 x-1

Question

Help!! How would I work this problem? And please show all your work :)

Answer 1

1+sin x = (2 cos^2 x)-1

cos²x+sin²x+sin x = (2 cos^2 x)-1,,,,,on left side 1=cos²x+sin²x

sin²x+sin x = cos²x-1=-sin²x

sinx(2sinx+1)=0

x=π,2π and x=7π/6,11π/6

Answer 2

If you meant 1+ sinx = [2cos^2(x)] -1

then

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

since sin^2(x) + cos^2(x) =1, replace cos^2(x) with 1-sin^2(x)

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

2sin^2(x) + sinx = 0

let y = sinx

2y^2 + y = 0

factor

y(2y+1) = 0

set each factor = 0

y=-1/2 or 0

sin^(x) = -1/2 or 0

x = sin^-1(-1/2) or sin^-1(0)

x = 7pi/6, 11pi/6, pi ot 2pi

BUT if you meant 1+sinx = 2cos^2(x-1)

then it has more complicated & messy calculations

odds are that's not how the problem was intended

but if you did or they did intend it that way,

there's a question whether the 1 in x-1 is 1 degree or 1 radian

assume it's 1 degree than x-1 approximately = x-0 = x

and x= about 3pi/2, pi/6 and 5pi/6

if 1 means 1 radian, then x-1 is about = x-pi/3

1+sin(x) = 2(1-sin^2(x-pi/3)

use sum and difference formula

sin(A+B) = sinAcosB+sinBcosA

1+sin(x) = 2 -2(sinxcos(-pi/3)+sin(-pi/3)cos(x))

-1+sinx = -2(sinx/2 + -sqr3/2)cosx)

-1+ sinx= -sinx - sqr3cosx

2sinx+sqr3cosx -1 = 0