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question mark

Find the general solution of
Sinx - 3sin2x + sin3x = cosx - 3cos2x + cos 3x
How can I get the answer to this? I did it by the sin A + sin B formula on both sides and then took out the common on both sides and then substracted the left side on the right...then too I am unable to get this

Shruti suresh jagdale , 7 Years ago
Grade 11
anser 1 Answers
Arun
sin(x)  –  3sin(2x) + sin(3x) = cos(x)  –  3cos(2x) + cos(3x) 
sin(2x - x) –  3sin(2x) + sin(2x + x) = cos(2x - x)  –  3cos(2x) + cos(2x + x) 
sin(2x)cos(x) - sin(x)cos(2x)  –  3sin(2x) + sin(2x)cos(x) + sin(x)cos(2x) = cos(2x)cos(x) + sin(2x)sin(x)  –  3cos(2x) + cos(2x)cos(x) - sin(2x)sin(x) 
2sin(2x)cos(x)  –  3sin(2x) = 2cos(2x)cos(x)  –  3cos(2x) 
sin(2x) * (2cos(x)  –  3) = cos(2x) * (2cos(x)  –  3) 
sin(2x) * (2cos(x)  –  3) - cos(2x) * (2cos(x)  –  3) = 0 
(sin(2x) - cos(2x)) * (2cos(x)  –  3) = 0 

2cos(x)  –  3 = 0 
2cos(x) = 3 
cos(x) = 3/2 
No solution 

sin(2x) - cos(2x) = 0 
sin(2x) = cos(2x) 
sin(2x)/cos(2x) = 1 
tan(2x) = 1 
2x = pi/4 + pi * k 
2x = (pi/4) * (1 + 4k) 
x = (pi/8) * (1 + 4k) 
 
Last Activity: 7 Years ago
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