How do you solve sin(x)-cos(x/3)=0?

2 Answers
May 17, 2018

x=(3pi)/8x=3π8

Explanation:

.

sinx-cos(x/3)=0sinxcos(x3)=0

sinx=cos(x/3)sinx=cos(x3)

We know:

sin(pi/2-theta)=costhetasin(π2θ)=cosθ

Therefore,

sin(pi/2-x/3)=cos(x/3)sin(π2x3)=cos(x3)

x=pi/2-x/3x=π2x3

x+x/3=pi/2x+x3=π2

6x+2x=3pi6x+2x=3π

8x=3pi8x=3π

x=(3pi)/8x=3π8

May 17, 2018

x = (3pi)/8 + (3kpi)/2x=3π8+3kπ2
x = (3pi)/4 + 3kpix=3π4+3kπ

Explanation:

sin x = cos (x/3)sinx=cos(x3)
cos (pi/2 - x) = cos (x/3)cos(π2x)=cos(x3)
Unit circle and property of cos function -->
pi/2 - x = +- x/3π2x=±x3
a. pi/2 - x = x/3π2x=x3
x + x/3 = pi/2x+x3=π2
(4x)/3 = pi/2 + 2kpi4x3=π2+2kπ
x = (3pi)/8 + (3kpi)/2x=3π8+3kπ2
b. pi/2 - x = - x/3π2x=x3
x - x/3 = pi/2xx3=π2
(2x/3) = pi /2 + 2kpi(2x3)=π2+2kπ
x = (3pi)/4 + 3kpix=3π4+3kπ