2cos(3x)=2+sin(3x)2cos(3x)=2+sin(3x), solve for xx (0<=x<=1800≤x≤180). solve this please?

1 Answer
Nghi N.
Oct 13, 2015

Solve 2cos 3x = 2 + sin 3x2cos3x=2+sin3x

Ans: 0^@19 and -17^@900∘19and−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 26.57tana=sinacosa=12=tan26.57 --> cos a = 0.89.
Equation (1) -->
cos 3x - (sin a/(cos a)) sin x = 1cos3x−(sinacosa)sinx=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(3x+a)=(3x+26,57)=±27∘13
a. 3x + 26.57 = 27.13 --> 3x = 0.56 --> x = 0.56/3 = 0^@19x=0.563=0∘19
b. 3x + 26.57 = - 27.13 --> x = - 53.70 --> x = -17^@90.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)(0∘19)

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