So we have to turn addition into multiplacation b/c only then can we use the rule that allows us to solve following problem (x-2)(x+3) = 0 so that x-2 =0 and x+ 3 = 0 gives x=2 and x=3. This law is called the ab=0 rule.
sin(6x)+ sin(2x) = 0 has a trig identity in sin(a) + sin(b) = 2sin( (a+b)/2)*sin((a-b)/2)
the above becomes 2sin((6x+2x)/2)sin((6x-2x)/2)=0
2sin(4x)sin(2x)=0 Now that our expression is multiplication (this is idea in other problems too through factoring or other methods turn problem into some form of multiplication of trig expression)
Set each factor = 0 sin(4x)= 0 solution to this comes from our knowledge of when sin(theta)=0
sin(theta) = 0 when theta = 0 + n pi (best to keep it general by using n so that we can answer all type of questions)
so theta is 4x so 4x = 0 + n pi now solve for x
x = 0/4 + n pi /4 Now let n cylce through its various values until > 2pi = 8pi/4 with our current denom.
x = (0, pi/4 , 2pi/4, 3pi/4, 4pi/4, 5pi/4, 6pi/4, 7pi/4, 8pi/4) last solution is striked because quest is between [0,2pi) where 0 is included but 2pi result is not. Notice also that I did not reduce the ans b/c these values are just easier to generate by not reducing first. I can reduce later
These are not the only solutions b/c sin(2x) = 0 as well
sin(theta) = 0 when theta = 0 + n pi thus 2x = 0 + n pi solve for x gives
x = 0/2 + n pi /2 again let n cycle through various values gives until > 2pi = 4pi/2 gives
x = 0, pi/2 , 3pi/2, 4pi/2 These are all stricken since they are just duplicates of previous ans after reducing.