<p>Algebraically, find all solutions for the equation sin(2x)=sin x for 0<=x<2pi. Give EXACT answers and show all work. </p>
<p>By the way, sin(2x)=2sin(x)cos(x).
So: 2sin(x)cos(x)=sin(x)</p>
<p>There's 4 solutions, I know how to figure out the ones by dividing both sides by sin(x), but I don't know how to get the the other 2: x=1/2, x=0.</p>
<p>Well THERE'S your problem. Man, I guess your teacher never taught you to never divide by sin x, cos x, or whatever.</p>
<p>sin 2x=sin x
sin x= sin 2x
sin x= 2sin x cos x
0= 2sin x cos x - sin x
0= sin x(2cos x -1)</p>
<p>sin x=0
x=0, pi</p>
<p>AND</p>
<p>2cos x - 1= 0
2cosx = 1
cos x = 1/2
x = pi/3 and 4pi/3</p>
<p>^^ i think you meant 5pi/3... 4pi/3 = -1/2</p>
<p>This forum should get LaTeX :D</p>
<p>Ah, yes. 4pi/3 is in the third quadrant. </p>
<p>x = 0, pi, pi/3 and 5pi/3.</p>
<p>AHHH GROSS! I just finished pre-calc last year.. and its like completely out of my head now... sin? cos? what's that? haha</p>
<p>I second that...although i don't know how to use it, LaTex would be very helpful.</p>