# How to solve the equation `sqrt(3)sin x-cosx=0` ?

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First, we'll shift cos x to the right side:

sqrt3*sin x = cos x

We'll divide by cos x both sides:

sqrt3*sin x/cos x = 1

But the fraction sin x/cos x can be replaced by the tangent function tan x.

sqrt3*tan x = 1

tan x = 1/sqrt3

Since it is not allowed to keep the square root to denominator, we'll multiply both, numerator and denominator, by sqrt3.

tan x = sqrt3/3

x = arctan (sqrt3/3) + k*pi

x = pi/6 + k*pi

**The set of solutions of the equation is: {pi/6 + k*pi}.**

sin x/cos x=1/sqrt(3)

tan x=1/sqrt 3

we know tan 30=1/sqrt 3

»x=30

The question is not very clear, i.e is it sqrt(3) sinx-cosx =0 or sqrt(3sinx)-cos x=0

I will do it for the questions:

1) sqrt(3) sinx - cos x =0

or sqrt (3) sinx = cos x

or, sinx/cosx = tanx = 1/sqrt(3)

i.e. **x = pi/6 (or 30 degrees)**

2) sqrt (3 sinx)-cosx= 0

or 3sinx = cos^2 x = 1- sin^2 x

(using sin^2 x + cos^2 x = 1)

or, sin^2x + 3 sinx -1 = 0

solving this quadratic equation, we get sin x = (1/2) x (-3+ sqrt (13))

or, **x = 17.62 degrees**.

`sqrt(3)*sinx - cosx = 0`

`or, (sqrt(3)/2)*sinx - (1/2)*cosx = 0`

`or, cos(30)*sinx - sin(30)*cosx = 0`

`or, sin(x-30) = 0`

`or, sin(x-30) = sin(n*pi)`

`or, ` `or, x-30 = n*pi`

`or, x = n*pi + 30`

`or, x = n*pi + pi/6`

``

`sqrt(3)sinx-cosx=0`

`sqrt(3)/2 sinx-(1/2)cosx=0`

`sin(pi/3) sinx-cos(pi/3)cosx=0`

`cos(x+pi/3)=cos(2n pi+pi/2)`

`x+pi/3=2n pi+pi/2`

`x=2n pi+(pi/2-pi/3)`

`x=2n pi+pi/6`

Ans.

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