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

Question

1529 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!

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