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

Question

42928 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-24T03:51:16

Let's first of all convert everything to #sinx#.

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

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

Now substituting into our equation:

#2(1 - sin^2x) - sinx - 1 = 0#

#2 - 2sin^2x - sinx - 1 =0 #

#-2sin^2x - sinx + 1 = 0#

#-2sin^2x - 2sinx + sinx + 1 = 0#

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

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

#sinx = 1/2 and sinx = -1#

#x = pi/6, (5pi)/6, (3pi)/2#

Note that these solutions are only in the interval #0 < x < 2pi#.

Hopefully this helps!