sin 2x - sin 3x = 0 ,find general solutions
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Sin2x - Sin3x =0
2Cos(5x/2)Sin(x/2) = 0
We can form 2 cases:
1) cos(5x/2) = 0
So, 5x/2 = (2n+1)pi/2
x = (2n+1)pi/5
2) sin(x/2) = 0
So, x/2 = npi
x= 2npi
So the general solutions are x = (2n+1)pi/5 or x = 2npi although the case of x = 2npi is already considered in x=(2n+1)pi/5 So we can say that the general solution is x = (2n+1)pi/5
Hope it helps
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2Cos(5x/2)Sin(x/2) = 0
We can form 2 cases:
1) cos(5x/2) = 0
So, 5x/2 = (2n+1)pi/2
x = (2n+1)pi/5
2) sin(x/2) = 0
So, x/2 = npi
x= 2npi
So the general solutions are x = (2n+1)pi/5 or x = 2npi although the case of x = 2npi is already considered in x=(2n+1)pi/5 So we can say that the general solution is x = (2n+1)pi/5
Hope it helps
Pls mark as brainliest
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