How do you solve 2sin(x-(pi/3))=sinx-squarerootof3cosx?

Question

This is a trigonometry identity problem. 

Answer 1

I am assuming the problem reads:

 

2sin(x - (pi/3)) = sinx - (sqrt(3))cosx, that is the square root of 3 is the coefficient of cosx and not that 3cosx is the radicand.

 

If this is the case, use the difference formula to simplify the left-hand side.

 

sin(A - B) = sinAsinB - cosAcosB

 

2[(sinx)sin(pi/3) - (cosx)cos(pi/3)] = sinx - (sqrt(3))cosx

 

pi/3 is the 60-degree angle in the 30-60-90 special right triangle.  Sine is sqrt(3)/2 and cosine is one half.

 

2[(sqrt(3)/2)sinx - (1/2)cosx] = sinx - (sqrt(3))cosx

 

Distribute the two on the left-hand side.

 

(sqrt(3))sinx - cosx = sinx - (sqrt(3))cosx

 

Collect like terms

 

[sqrt(3) - 1]sinx = [1 - sqrt(3)]cosx

 

Divide through by cosx

 

[sqrt(3) - 1]tanx = [1 - sqrt(3)]

 

Factor negative one out of right-hand side.

 

[sqrt(3) - 1]tanx = - [sqrt(3) - 1]

 

Divide through by radical expression.

 

tanx = - 1

 

Tangent is negative in the second and fourth quadrants.

 

x = 3pi/4, 7pi/4, 11pi/4, . . . ., (4n - 1)pi/4 for n = 1 to infinity.

 

For the negative values:  x = - pi/4, - 5pi/4, -9pi/4, ... -(n + 4)pi/4 for n = 0 to infinity.