Question

Solve algebraically cos2xsinx+sinx=0 with solutions in the interval [0,2pi)?

Been stuck on this homework question for a bit, any help or explanation? I you convert cos2x to (1+2sin(x)) just don't know where to go from there

Answer 1

`0, pi/2, pi, (3pi)/2, 2pi`

Explanation:

f(x) = cos 2x.sin x + sin x = 0
f(x) = sin x(cos 2x + 1) = 0
Either factor of the product should be zero:
a. sin x = 0
Unit circle gives -->
x = 0, `x= pi, and x = 2pi`
b. cos 2x = 1 = 0 --> cos 2x = -1
`2x = +- pi`
`x = +- pi/2`, or
`x = pi/2`, and `x = (3pi)/2` (co-terminal)

Answer 2

`x = 0, pi, pi/2, (3pi)/2`

Explanation:

Start by recalling that `cos(2x) = 1 - 2sin^2x`.

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

`sinx - 2sin^3x + sinx = 0`

`2sinx - 2sin^3x = 0`

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

`sinx = 0 and sinx = +- 1`

`x = 0, pi, pi/2, (3pi)/2`

A graph confirms:

Hopefully this helps!

Answer 3

The solutions are `x=0,pi/2,pi,(3pi)/2`.

Explanation:

We need to use one of the three cosine double-angle formulae:

`cos(2x)=2cos^2x-1`

`color(white)cos(2x)=cos^2x-sin^2x`

`color(white)cos(2x)=1-2sin^2x`

In this case, I'll choose the third one so that the equation will only have sines in it.

Here's what it will look like:

`cos(2x)sinx+sinx=0`

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

`sinx-2sin^3x+sinx=0`

`2sinx-2sin^3x=0`

`2sinx-2(sinx)^3=0`

Now, we can use a substitution, then solve the equation like a regular polynomial.

Let `u=sinx`:

`2u-2u^3=0`

`2u(1-u^2)=0`

`u(1-u^2)=0`

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

`u=-1,0,1`

Now, plug `sinx` back in for `u`:

`sinx=-1,0,1`

Here's a unit circle to remind us of some sine values:

We can see that `sinx` is `-1` when `x` is `(3pi)/2`, it's `0` when `x` is `0` or `pi`, and it's `1` when `x` is `pi/2`. That means that all the solutions together are:

`x=0,pi/2,pi,(3pi)/2`

That's it. Hope this helped!