Question

Solve the equation `cos^2x-sin^2x=sinx+1` in the interval `(0<=x<=2pi)`?

Answer 1

`x={0,(5pi)/6,pi,(7pi)/6,2pi}`

Explanation:

As `cos^2x-sin^2x=sinx+1`

we have `1-sin^2x-sin^2x=sinx+1`

or `2sin^2x+sinx=0`

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

i.e. either `sinx=0=sin0` i.e. `x=npi`

or `sinx=-1/2=sin(-pi/6)` i.e. `x=npi+(-1)^n(-pi/6)`

and with in period `[0,2pi]`, we have

`x={0,(5pi)/6,pi,(7pi)/6,2pi}`

Answer 2

`x=0, (5pi)/6, pi, (7pi)/6, 2pi` `(0<=x<=2pi)`

Explanation:

`cos^2x-sin^2x=sinx+1`
`(1-sin^2x)-sin^2x=sinx+1`
`-2sin^2x cancel(+1)=sinx cancel(+1)`
`2sin^2x+sinx=0`
`sinx(2sinx+1)=0`

Here we the above equation to get two new equations (I'm assuming you know how to solve quadratic equations):

`sinx=0`
`therefore x=0, pi, 2pi` `(0<=x<=2pi)`

OR

`2sinx+1=0`
`sinx=-1/2`

Here, we cannot go straight to saying `x=pi/6`, as this would give a positive answer. Sine is negative in the 3rd and 4th quadrants, so these of `x` in `sinx=-1/2` can be found by putting the acute positive angle `x=pi/6` into `pi+x`, and `2pi-x`.

So,

`2sinx+1=0`
`sinx=-1/2`
`x= pi+pi/6, 2pi-pi/6`
`therefore x=(5pi)/6, (7pi)/6` `(0<=x<=2pi)`

Putting these two sets of answers together:
`therefore x=0, (5pi)/6, pi, (7pi)/6, 2pi` `(0<=x<=2pi)`

However this is assuming that the domain the question asked for is inclusive of `0` and `2pi`.

If it is exclusive of these, the solutions to this equation will be:
`x= (5pi)/6, pi, (7pi)/6` `(0 < x < 2pi)`

Solving this is just like solving an algebraic equation once you get everything in terms of `sinx`, so be sure to know your algebra and trig identities back to front!