Why does sin(x)=0 have solutions 0, π, 2π and cos(x)=0 have solutions π/2, 3π/2?

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Just a quick question, I was solving sin(x)=0 and cos(x)=0. I was trying hard to find out the solutions and the solutions were: for sin(x)=0------>x=0, pi, 2pi for cos(x)=0-------->x=pi/2, 3pi/2

But my question is why are those the solutions?
 
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Ric-Veda said:
Just a quick question, I was solving sin(x)=0 and cos(x)=0. I was trying hard to find out the solutions and the solutions were: for sin(x)=0------>x=0, pi, 2pi for cos(x)=0-------->x=pi/2, 3pi/2

But my question is why are those the solutions?

There are more solutions! (sin and cos are periodic functions and repeat itself every ##2\pi##)
 
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Math_QED said:
There are more solutions! (sin and cos are periodic functions and repeat itself every ##2\pi##
But I want to know the whole step. Do I get the values by just looking at their domains from their graphs?
 
Thread moved from Precalc Homework section. @Ric-Veda, if you post in the Homework & Coursework sections, like you originally did with this thread, you must use the homework template.

Since you've posted other questions about Laplace transforms and differential equations, it's a reasonable expectation that you have some familiarity with trig functions. If not, you really need to spend some time reviewing them.

As already mentioned, the sine and cosine functions are periodic. Take a look at the graphs of these functions, which should make it obvious that there are many (infinitely many) solutions to sin(x) = 0 and cos(x) = 0.
 
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Ric-Veda said:
But I want to know the whole step. Do I get the values by just looking at their domains from their graphs?
Try drawing a triangle and imagine the angle formed when the adjacent or opposite side is 0.