When I solve the equation sin2(x)=sin(x), do I isolate the number first and then try to solve?

Question

It just says to solve the equation...so I'm not quite sure where to start.

Its actually sin(2x)=sin(x)

Answer 1

Use the trig identity sin(2x) = 2sin(x)cos(x)

So your problem is:

2sin(x)cos(x) = sin(x) Divide by sin(x) gives

2cos(x) = 1 so cos(x) = 1/2

x = 60 degrees or 300 degrees (plus multiples of 360)

or if you wish x = pi/3 or 5pi/3 (plus multiples of 2pi)

Here's some trig identities that are quite helpful:

http://www.sosmath.com/trig/Trig5/trig5/trig5.html

Answer 2

If that's supposed to be sin^2(x), then the way you solve it is to get all the terms on one side and then factor it like a quadratic. Then you'll have 2 equations to solve of the form sin(x) = something. Then you use inverse sine to find the values of x that make the equation true.

Answer 3

(sin(x))^2 = sin(x)

sin(x)^2 - sinx = 0
(sin(x))(sin(x) - 1) = 0

sin(x) = 0
sin(x) - 1 = 0

sin(x) = 0
x = 0°, 180°, 360°

sin(x) - 1 = 0
sin(x) = 1
x = 90°

so

x = 0°, 90°, 180°, 360°

Answer 4

you isolate what your solving for be it the # or the x or the sin

Answer 5

sin(x) = sin (2x)
sin(x) = 2cos(x)sin(x)
sin(x) - 2cos(x)sin(x) = 0
sin(x)(1- 2cos(x)) = 0
sin(x) = 0 or 1 - 2cos(x) = 0 => cos(x) = 0.5
x = Pi*n or x = Pi/3 + 2Pi*n or x = 2Pi/3 + 2Pi*n

Answer 6

Yeah,
if sin(2x)=sin(x), then 2x=x+n*pi, where n=...-2,-1,0,1,2...
so,x=n*pi.