Question

Solve: 2cos(x)-square root 3=0 for 0 less than x less than 2 pi

Answer 1

Answer:

The general solution of   $cos x = cos(\frac{\pi }{6})$   is  

                                                x = 2nπ±$\frac{\pi }{6}$

The general solution values  

                                 $x = - \frac{\pi }{6} and x = \frac{\pi }{6}$

Step-by-step explanation:

Explanation:-

Given equation is  

                              $2cosx-\sqrt{3} =0 for 0$

                              $2cosx =\sqrt{3}$

Dividing '2' on both sides, we get

                             $cos x =\frac{\sqrt{3} }{2}$

                             $cos x = cos(\frac{\pi }{6})$

General solution of cos θ = cos ∝ is θ = 2nπ±∝

Now The general solution of   $cos x = cos(\frac{\pi }{6})$   is  

                                                x = 2nπ±$\frac{\pi }{6}$

put n=0

$x = - \frac{\pi }{6} and x = \frac{\pi }{6}$

Put n=1  

$x = 2\pi +\frac{\pi }{6} = \frac{13\pi }{6}$

$x = 2\pi -\frac{\pi }{6} = \frac{11\pi }{6}$

put n=2

$x = 4\pi +\frac{\pi }{6} = \frac{25\pi }{6}$

$x = 4\pi -\frac{\pi }{6} = \frac{23\pi }{6}$

And so on

But given 0 < x< 2π

The general solution values  

                                 $x = - \frac{\pi }{6} and x = \frac{\pi }{6}$