Trigonometric equations - strange results

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The discussion revolves around solving the trigonometric equation 2sin(2A) = tan(A) and the resulting confusion over missing angles. The original poster successfully calculated four angles but is unsure how to find the others. A key point raised is the mistake of dividing by sin(A) without considering the possibility that sin(A) could be zero, which leads to missing solutions. The correct approach involves factoring the equation to account for both cases: when sin(A) equals zero and when it does not. This highlights the importance of not dividing by terms that may equal zero in trigonometric equations.
repugno
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Greetings all,

I am getting strange results when solving this trig equation. I seem to be able to calculate 4 out 7 of the correct angles but how do i calculate the others? Maybe my method is wrong...

A = theta

2sin2A = tanA,

considering identities sin2A = 2sinAcosA and tanA = sinA/cosA

2(2sinAcosA) = sinA/cosA

4sinAcos^2A = sinA

dividing by sinA both sides

4cos^2A = 1
cos^2A = 1/4
cosA = 1/2
cosA = -1/2

A = 60
A = 300
A = 120
A = 240

missing angles 0, 180 and 360 ?? :confused:

Any help would be much appreciated, thanks
 
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When dividing with sin(A), you have assumed sin(A) not equal to zero.
 
I think that you have to illustrate that

4sin(A)cos^2(A) = sin(A)

4sin(A)cos^2(A) - sin(A) = 0

So

sin(A)(4cos^2(A) - 1) = 0
 
Or, alternatively, you have to break the problem up into two cases; case 1 is where sin A is 0, and case 2 is where sin A is not zero (and thus you can divide by sin A)


But whatever you do, the point we're making is that, in general, you cannot divide by something that may be zero in your problem.
 
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