Question

Question #fce84

Answer 1

`x=pi/2+kpi,pi/4+(kpi)/2,k in ZZ`

Explanation:

Use the identity `sin2x=2sinxcosx` to rewrite the expression.

`sin2xsinx=cosx" "=>" "2sinxcosx(sinx)=cosx`

`=>" "2sin^2xcosx=cosx`

Subtract `cosx` from each side.

`2sin^2xcosx-cosx=0`

Factor `cosx`.

`cosx(2sin^2x-1)=0`

Set both of these terms equal to `0`.

`cosx=0`

This occurs at `x=pi/2` and `x=(3pi)/2` on the interval `[0,2pi)`, but can be generalized as `x=pi/2+kpi` where `k` is an integer.

The other term is

`2sin^2x-1=0`

`sin^2x=1/2`

`sinx=+-sqrt(1/2)=+-1/sqrt2=+-sqrt2/2`

Sine equals `+-sqrt2/2` at `x=pi/4,(3pi)/4,(5pi)/4,(7pi)/4` on `[0,2pi)`, which can be generalized to `pi/4+(kpi)/2`, where `k` is an integer.

Note that the mathematical way to express that `k` is an integer is to say that `k in ZZ`.