Hey! Okay so i don't seem to be able to find an easy way to do this (like I got an answer but the method was waaay too complicated I reckon but I got -8 degrees and 48 seconds and -24 degrees and 54 seconds??) D: I was wondering whether you guys could help me! thanks
Solve the following equation:
2cos2ϴ=1-3sin2ϴ 0≤ϴ≤360
Thank you!
Neutron
Hi Neutron,
I think, I found an easy answer for your question, 2cos(2x) = 1-3sin(2x).Before I prove it, I need to point out that cos(x) = 0 is not a solution of the equation since that forces LHS=-2(because cos(2x) =2cos^2(x) -1 ) and RHS = 1(since sin(2x) = 2 sin(x) cos(x))
Solution:
2cos(2x) = 1-3sin(2x)
3(cos(2x) + sin(2x)) = 1+cos(2x) = 2cos^2(x)
3(cos^2(x) -sin^2(x) +2sin(x)cos(x) ) = 2cos^2(x) (since cos(2x) = cos^2(x) -sin^2(x) )
collecting cos^2(x) terms yields
cos^2(x) = sin(x) (3sin(x) - 6cos(x))
1 = (sin(x) / cos(x)) ( 3sin(x) -6cos(x) )/cos(x) (dividing by cos^2(x), it is possible since cos(x) is not zero )
1 = tan(x)(3tan(x) - 6)
3tan^2(x) -6tan(x) -1 = 0
tan(x) = 1+2sqrt(3)/3 or 1-2sqrt(3)/3
From there you can find the angles.I do agree with -8 degree but I don't think -24 degree is correct, I put it into the calculator and it gave me
RHS= 1.33 , LHS =3.23
A general tip to tackle to these type of questions where you get ' sin(x)cos(x)' somewhere when you solve it is to represented in terms of tan(x), (If you get lucky! )
I hope that was helpful. (Y)