How do you solve #3sin theta - 4 cos theta = 2#?

2 Answers
Apr 18, 2016

#S={209.551...+360^@n, 76.708...+360^@n}#

Explanation:

Use the following formulas to transform the equation then solve
#A cos x + B sin x = C cos (x – D)#

#C = sqrt(A^2+B^2 ), cos D = A/C and sin D = B/C#

#A=-4, B=3,C=sqrt(16+9)=sqrt25=5#

#cosD=-4/5->D=cos^-1 (-4/5) =143.13...^@#-> Quadrant II

#5cos(theta-143.13...^@)=2#

#cos(theta-143.13...^@)=2/5#

#theta-143.13...^@=cos^-1(2/5)#

#theta-143.13...^@=+-66.4218...+360^@n#

#theta=143.13...+-66.4218...+360^@n#

#theta = 209.551...+360^@n, 76.708...+360^@n#

#S={209.551...+360^@n, 76.708...+360^@n}#

Apr 18, 2016

#x = 76^@71 + 2kpi#
#x = 209.55 + 2kpi#

Explanation:

Divide both sides by 3 -->
#sin x - (4/3)cos x = 2/3#
Call #tan u = (sin u)/(cos u) = 4/3 #--> #u = 53^@13#
#sin x.cos u - sin u.cos x = (2/3)cos u = (2/3)0.60 = 0.40#
Apply the trig identity: sin (a - b) = sin a.cos b - sin b.cos a -->
sin (x - 53.13) = 0.40 --> sin (23.58) and sin (180 - 23.58)
Calculator and unit circle give 2 solution arcs:

#a. (x - 53.13) = 23^@58# --> #x = 23.58 + 53.13 = 76^@71#
#b. (x - 53.13) = 180 - 23.58 = 156^@42 #--> x = 156.42 + 53.13 =
#= 209^@55#