Question

19. Solve sin O+ 1 = cos 20 on the interval 0≤x < 2xpi

Answer 1

`\theta=(\pi)/(2),(3\pi)/(2),(2\pi)/(3),(4\pi)/(3)`

Explanation:

If I'm interpreting that correctly, you are trying to solve this equation:

`sin(\theta )+1=cos(2\theta)`

for theta. To do this, you will need a trig identity sheet (I'm assuming you got one from class) and a unit circle (ditto on the class thing).

We need to solve for theta. If I look to my trig identities, I will see a double angle one there that says:

`cos(2\theta)=1-2sin^2(\theta)`

We will make that replacement, then we will have everything in terms of sin.

`sin(\theta)+1=1-2sin^2(\theta)`

Now get everything on one side of the equals sign to solve for theta:

`2sin^2(\theta)+sin(\theta)=0`

We can factor out the common sin(theta):

`sin\theta(2sin\theta+1)=0`

By the Zero Product Property, either

`sin\theta=0` or

`2sin\theta+1=0`

Now look at your unit circle and find that the values of theta where the sin is 0 are located at:

`\theta=(\pi )/(2),(3\pi)/(2)`

The next one we have to solve for theta:

`2sin\theta+1=0` simplifies to

`2sin\theta=-1` and

`sin\theta=-(1)/(2)`

Look at the unit circle again to find the values of theta where the sin is -1/2:

`\theta=(2\pi)/(3),(4\pi)/(3)`

Those ar your values of theta!