How do you solve #cos^4x-sin^4x=sinx#?

Question

1201 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-24T04:02:16

#(cos^2x - sin^2x)(cos^2x + sin^2x) = sinx#

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

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

#1 - sin^2x - sin^2x - sinx = 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 and (3pi)/2#

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

Hopefully this helps!