Finding out degrees when solving trig equations

[coasterguy]

Ok, I have this:

Solve:
Sin? = sin2?

So...

sin?-2sin?cos? = 0

sin(-2cos?+1)=0

sin = 0 or 1-2cos?=0

The answer is "0degrees, 60 degrees, 180 degrees, 300 degrees".

But how did they get that answer???
The same thing goes for other questions, say:

sin2?=4 / sqrt3
The answer is "30degrees, 60 degrees, 210 degrees, 240 degrees". Why? Thanks

 

[wjm11]

Sin? = sin2?

So...

sin?-2sin?cos? = 0

sin(-2cos?+1)=0

sin = 0 or 1-2cos?=0

The answer is "0degrees, 60 degrees, 180 degrees, 300 degrees".

But how did they get that answer???
The same thing goes for other questions, say:

sin2?=4 / sqrt3
The answer is "30degrees, 60 degrees, 210 degrees, 240 degrees". Why? Thanks

First of all, I’m sure the problem states that you should give answers between 0 and 360. Second, you must consider answers in all four quadrants.

Now consider: how many times does sine theta equal zero between 0 and 360? Your calculator will only give you one of the answers. How many times does cosine theta equal 1/2? Again, your calculator will only give you one answer.

To see what is going on, I recommend you graph it. If you have a graphing calculator, enter y = sin?-2sin?cos? and look at the graph.

How many times does this cross the x-axis between 0 and 360? Those points are all solutions.

 

[galactus]

A good calculator will give you:

\(\displaystyle x=2C\pi+\frac{\pi}{3}, \;\ x=2C{\pi}-\frac{\pi}{3}, \;\ x=2C{\pi}\)



Which, as wjm pointed out, is due to the periodicity of the function.

 

[fasteddie65]

sin 2? = 4/?3

Are you sure you copied this problem correctly? As it stands now, there are no solutions.