#2cos(3x)=2+sin(3x)#, solve for #x# (#0<=x<=180#). solve this please?
1 Answer
Oct 13, 2015
Solve
Ans:
Explanation:
2cos 3x - sin 3x = 2. Divide both sides by 2.
cos 3x - (1/2)sin 3x = 1 (1)
Call
Equation (1) -->
cos 3x.cos a - sin a.sin 3x = cos a
cos (3x + a) = cos a = 0.89
a. 3x + 26.57 = 27.13 --> 3x = 0.56 -->
b. 3x + 26.57 = - 27.13 --> x = - 53.70 -->
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