Question

Find all solutions in the interval from [0,2pi) 2cos(3x)= -sqrt{2}

Answer 1

Solutions of $2cos(3x)= -\sqrt{2}$ in the interval from [0,2pi) is $x =\frac{\pi}{12}$  and $x = \frac{23\pi}{12}$ .

Step-by-step explanation:

Find all solutions in the interval from [0,2pi)

$2cos(3x)= -\sqrt{2}$

⇒ $2cos(3x)= -\sqrt{2}$

⇒ $\frac{2cos(3x)}{2}= \frac{-\sqrt{2}}{2}$

⇒ $cos3x= \frac{-\sqrt{2}(\sqrt{2})}{2{\sqrt{2}}}$

⇒ $cos3x= \frac{-2}{2{\sqrt{2}}}$

⇒ $cos3x= \frac{-1}{{\sqrt{2}}}$

⇒ $cos^{-1}(cos3x)= cos^{-1}(\frac{-1}{{\sqrt{2}}})$

⇒ $3x=\pm \frac{\pi}{4}$

⇒ $x=\pm \frac{\pi}{12}$

Cosine General solution is :

$x = \pm cos^{-1}(y)+ 2k\pi$

⇒ $x = \pm \frac{\pi}{12}+ 2k\pi$ , k is any integer .

At k=0,

⇒ $x =\frac{\pi}{12}$ ,

At k=1,

⇒ $x = - \frac{\pi}{12}+ 2\pi$

⇒ $x = \frac{23\pi}{12}$

Therefore , Solutions of $2cos(3x)= -\sqrt{2}$ in the interval from [0,2pi) is $x =\frac{\pi}{12}$  and $x = \frac{23\pi}{12}$ .