Question

How do you simplify `sin2x + cosx = 0` using the double angle identity?

Answer 1

General solution is `x=(2n+1)pi/2` or `x=npi+-(-1)^n(-pi/6)`, where `n` is an integer.
In the interval `[0,2pi)`, `x in{pi/2,(7pi)/6,(3p)/2,(11pi)/6}`

Explanation:

As `sin2x=2sinxcosx`, we can write `sin2x+cosx=0` as

`2sinxcosx+cosx=0`

or `cosx(2sinx+1)=0`

Hence either `cosx=0` or `2sinx+1=0` i.e. `sinx=-1/2`

If `cosx=0`, `x=(2n+1)pi/2`, where `n` is an integer

and if `sinx=-1/2=sin(-pi/6)`, `x=npi+-(-1)^n(-pi/6)`

in the interval `[0,2pi)`,

`x in{pi/2,(7pi)/6,(3p)/2,(11pi)/6}`