+23245 tan(2x) = [ 2tan(x) ] / [ 1 - tan2(x) ]
Problem: tan(2x) = sin(x)
---> [ 2·tan(x) ] / [ 1 - tan2(x) ] = sin(x)
2·tan(x) = sin(x)· [ 1 - tan2(x) ]
2·sin(x) / cos(x) = sin(x)· [ 1 - tan2(x) ]
2 / cos(x) = 1 - tan2(x) ( divide both sides by sin(x); now you must check for sin(x) = 0 )
2 / cos(x) = 1 - sin2(x) / cos2(x) ( replace tan with sin/cos )
2·cos(x) = cos2(x) - sin2(x) ( multiply both sides by cos2(x ))
2·cos(x) = cos2(x) - ( 1 - cos2(x) ) ( sin2(x) = 1 - cos2(x) )
2·cos(x) = cos2(x) - 1 + cos2(x)
0 = 2·cos2(x) - 2cos(x) - 1
Quadratic formula: cos(x) = [ 2 + sqrt(12) ] / 4 or cos(x) = [ 2 + sqrt(12) ] / 4
So: possible answers: x = sin-1(0) x = cos-1( [ 2 + sqrt(12) ] / 4 ) x = cos-1( [ 2 - sqrt(12) ] / 4 )
You'll need to check these to see which one or ones are correct solutions.
