SOLUTION: what are the solutions to sin2x=sinx

Question 373466: what are the solutions to sin2x=sinx
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
sin(2x) = sin(x)
Using the identity sin(2x) = 2sin(x)cos(x) this becomes:
2sin(x)cos(x) = sin(x)
Subtracting sin(x) from each side:
2sin(x)cos(x) - sin(x) = 0
Factoring out sin(x):
sin(x)(2cos(x) - 1) = 0
Using the Zero Product property:
sin(x) = 0 or 2cos(x) - 1 = 0
Solving the second equation for cos(x) we get:
sin(x) = 0 or cos(x) = 1/2
So our solutions are all the angles whose sin is 0 or all the angles whose cos is 1/2. Since "x" was used for the angle instead of theta, I'm assuming we want radian measure angles. (If you need angles in degrees, just replace all the 's in the answers we get with 180 and simplify.)

Sin is zero at 0 and and for all coterminal angles. So for the first equation above, the solution is expressed as:
x = 0 + n (where n is any integer)
or
x = + n (where n is any integer)
(Note: The "+ n (where n is any integer)" is how we express the idea: "and for all coterminal angles".)

For the second equation, cos is 1/2 for and and all coterminal angles. So the solution for the second equation is:
x = + n (where n is any integer)
or
x = + n (where n is any integer)

The complete solution is:
x = 0 + n (where n is any integer)
or
x = + n (where n is any integer)
or
x = + n (where n is any integer)
or
x = + n (where n is any integer)

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