You should come up with the substitution angle = `alpha` , hence:

`sin^4 alpha + cos^4 alpha = 3/4`

You need to use the basic formula of trigonometry such that:

`sin^2 alpha + cos^2 alpha = 1`

Raising to square both sides yields:

`(sin^2 alpha + cos^2 alpha)^2 = 1 =gt sin^4 alpha + cos^4 alpha + 2sin^2 alpha*cos^2 alpha = 1`

The problem provides the information `sin^4 alpha + cos^4 alpha = 3/4` , hence you need to substitute `3/4` for `sin^4 alpha + cos^4 alpha` in`sin^4 alpha + cos^4 alpha + 2sin^2 alpha*cos^2 alpha = 1` such that:

`3/4 + 2sin^2 alpha*cos^2 alpha = 1 =gt 2sin^2 alpha*cos^2 alpha = 1 - 3/4`

`` `2sin^2 alpha*cos^2 alpha = 1/4 =gt sin^2 alpha*cos^2 alpha=1/8`

You should consider the following possibilities for `sin^2 alpha*cos^2 alpha=1/8` such that:

`sin^2 alpha = 2/4 and cos^2 alpha=1/4 =gt sin^2 alpha*cos^2 alpha=(2/4)*(1/4) = 2/16 = 1/8`

`sin^2 alpha = 1/4 and cos^2 alpha=2/4 =gt sin^2 alpha*cos^2 alpha=(1/4)*(2/4) = 2/16 = 1/8`

Hence, if `sin^2 alpha = 2/4 =gt sin alpha = +- sqrt 2/2 =gt alpha = +- pi/4` .

If `sin^2 alpha = 1/4 =gt sin alpha = +- 1/2 =gt alpha = +- pi/6` If `cos^2 alpha=1/4 =gtcos alpha = +- 1/2 =gt alpha = +- pi/3`

**Hence, evaluating the solutions to the equation yields: `alpha = +-pi/6 ; alpha = +-pi/4 ; alpha = +-pi/3` .**