#2cos(3x)=2+sin(3x)#, solve for #x# (#0<=x<=180#). solve this please?

1 Answer
Oct 13, 2015

Solve #2cos 3x = 2 + sin 3x#

Ans: #0^@19 and -17^@90#

Explanation:

2cos 3x - sin 3x = 2. Divide both sides by 2.
cos 3x - (1/2)sin 3x = 1 (1)
Call #tan a = sin a/(cos a) = 1/2 = tan 26.57# --> cos a = 0.89.
Equation (1) -->
#cos 3x - (sin a/(cos a)) sin x = 1#
cos 3x.cos a - sin a.sin 3x = cos a
cos (3x + a) = cos a = 0.89
#(3x + a) = (3x + 26,57) = +- 27^@13#
a. 3x + 26.57 = 27.13 --> 3x = 0.56 --> #x = 0.56/3 = 0^@19#
b. 3x + 26.57 = - 27.13 --> x = - 53.70 --> #x = -17^@90.#
Check by calculator.
x = -17.90 --> 3x = -53.70 --> 2cos 3x =1.18 --> sin 3x = - 0.8
2cos 3x = 2 =sin 3x --> 2.18 = 2 - 0.80. OK
x = 0.19 --> 3x = 0.56 --> 2cos 3x = 2.00 --> sin 3x = 0.01
1.99 = 2 + 0.1 . OK
NOTE Within interval (0, 180), there is one answer #(0^@19)#