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marusya05 [52]
3 years ago
10

Solve sin 0 + 1 = cos20 on the interval 0 ≤ 0 < 2pi. Show work please!

Mathematics
1 answer:
yan [13]3 years ago
7 0

Answer:

\theta=\frac{\pi}{2},\frac{3\pi}{2}\frac{2\pi}{3}\frac{4\pi}{3}

Step-by-step explanation:

You need 2 things in order to solve this equation:  a trig identity sheet and a unit circle.

You will find when you look on your trig identity sheet that

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

so we will make that replacement, getting everything in terms of sin:

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

Now we will get everything on one side of the equals sign, set it equal to 0, and solve it:

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

We can factor out the sin(theta), since it's common in both terms:

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

Because of the Zero Product Property, either

sin(\theta)=0 or

2sin(\theta)+1=0

Look at the unit circle and find which values of theta have a sin ratio of 0 in the interval from 0 to 2pi.  They are:

\theta=\frac{\pi}{2},\frac{3\pi}{2}

The next equation needs to first be solved for sin(theta):

2sin(\theta)+1=0 so

2sin(\theta)=-1 and

sin(\theta)=-\frac{1}{2}

Go back to your unit circle and find the values of theta where the sin is -1/2 in the interval.  They are:

\theta=\frac{2\pi}{3},\frac{4\pi}{3}

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