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depends on the trig but should be able to
it's solving equations using identities.
yea depends :3
Find all values of (theda symbol) on the interval 0 more than or equal to (theda) more than or equal to 360 that solve sin(2(theda))=sin(theda)
Can't type symbols lol
Find all values of Ө on the interval 0≤Ө≤360 that solve sin(2Ө)=sinӨ
ok this is fairly simple
you know that sine is a function who's curve repeats it self over and over, a function that does this is called a periodic function, meaning it has a period upon which it repeats its self, in the case of sine and cosine they are periodic about 2PI
sine at zero is equal zero, one could also say that sin(0*Pi) = 0 because zero times pi also equals zero thus it is the same as saying sin(0)=0
the sine curve oscillates between 1 and -1 with a period of 2 pi, so at zero sin =0, at 90* which is equal to pi/2 radians sine hits one, then comes back down to zero at 180* or pi radians, it then goes down under the x axis and touches -1 at sine(270*) or 3pi/2 radians, then back up to zero when it hits 360* or 2pi radians. so sine(0)=0 , sine(pi/2)=1, sine(pi)=0, sine(3pi/2) = -1, and sine(pi)=0
however there are other values between these angles such as sin(pi/4) , pi/6, ect... so you would need to check for those values as well. the easiest way to do this is by drawing or looking at a unit circle. are you familiar with the unit circle?