Question

Solve the equation `cos2theta+sintheta=0`, if `0<=theta<=2pi`?

Answer 1

`theta=pi/2,(7pi)/6, (11pi)/6`

Explanation:

`cos2θ + sinθ = 0`
`cos(2theta)=1-2sin^2theta`
`:. cos2θ + sinθ=1-2sin^2theta+sintheta=0`
(rearranging to look more like a quadratic)
`-2sin^2theta+sintheta+1=0`
`color(red)(-1)(-2sin^2theta+sintheta+1)=color(red)(-1)xx0`
`2sin^2theta-sintheta-1=0`
Let `x=sintheta`
`2x^2-x-1=0`
Factorising the quadratic yields;
`(2x+1)(x-1)=0`
By the null factor law;
`x=-1/2` or `x=1`
`:. sintheta=-1/2` or `sintheta=1`
Hence,
`theta=pi/2,(7pi)/6, (11pi)/6` on the domain `[0,2pi]`

I hope that helps :)

Answer 2

`theta=pi/2` or `(7pi)/6` or `(11pi)/6`

Explanation:

We have

`cos2theta+sintheta=0`, solutions in `0<=theta<=2pi`

Substitute `cos^2theta=1-2sin^2theta`

`1-2sin^2theta+sintheta=0`

Let `sintheta=u`

`1-2u^2+u=0`

`<=>2u^2-u-1=0`

or `(2u+1)(u-1)=0`

`u=1, u=-1/2`

Substitute back `sintheta=1` or `-1/2`

and for `0<=theta<=2pi`

`theta=pi/2` or `pi+pi/6=(7pi)/6` or `2pi-pi/6=(11pi)/6`