How do you solve 2cos^2x-sinx-1=02cos2xsinx1=0?

1 Answer
Jul 24, 2016

Let's first of all convert everything to sinxsinx.

Consider the pythagorean identity sin^2x + cos^2x = 1sin2x+cos2x=1. Rearranging:

cos^2x = 1 - sin^2xcos2x=1sin2x

Now substituting into our equation:

2(1 - sin^2x) - sinx - 1 = 02(1sin2x)sinx1=0

2 - 2sin^2x - sinx - 1 =0 22sin2xsinx1=0

-2sin^2x - sinx + 1 = 02sin2xsinx+1=0

-2sin^2x - 2sinx + sinx + 1 = 02sin2x2sinx+sinx+1=0

-2sinx(sinx + 1) + 1(sinx + 1) = 02sinx(sinx+1)+1(sinx+1)=0

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

sinx = 1/2 and sinx = -1sinx=12andsinx=1

x = pi/6, (5pi)/6, (3pi)/2x=π6,5π6,3π2

Note that these solutions are only in the interval 0 < x < 2pi0<x<2π.

Hopefully this helps!