Solutions of [tex]2cos(3x)= -\sqrt{2}[/tex] in the interval from [0,2pi) is [tex]x =\frac{\pi}{12}[/tex] and [tex]x = \frac{23\pi}{12}[/tex] .
Step-by-step explanation:
Find all solutions in the interval from [0,2pi)
[tex]2cos(3x)= -\sqrt{2}[/tex]
⇒ [tex]2cos(3x)= -\sqrt{2}[/tex]
⇒ [tex]\frac{2cos(3x)}{2}= \frac{-\sqrt{2}}{2}[/tex]
⇒ [tex]cos3x= \frac{-\sqrt{2}(\sqrt{2})}{2{\sqrt{2}}}[/tex]
⇒ [tex]cos3x= \frac{-2}{2{\sqrt{2}}}[/tex]
⇒ [tex]cos3x= \frac{-1}{{\sqrt{2}}}[/tex]
⇒ [tex]cos^{-1}(cos3x)= cos^{-1}(\frac{-1}{{\sqrt{2}}})[/tex]
⇒ [tex]3x=\pm \frac{\pi}{4}[/tex]
⇒ [tex]x=\pm \frac{\pi}{12}[/tex]
Cosine General solution is :
[tex]x = \pm cos^{-1}(y)+ 2k\pi[/tex]
⇒ [tex]x = \pm \frac{\pi}{12}+ 2k\pi[/tex] , k is any integer .
At k=0,
⇒ [tex]x =\frac{\pi}{12}[/tex] ,
At k=1,
⇒ [tex]x = - \frac{\pi}{12}+ 2\pi[/tex]
⇒ [tex]x = \frac{23\pi}{12}[/tex]
Therefore , Solutions of [tex]2cos(3x)= -\sqrt{2}[/tex] in the interval from [0,2pi) is [tex]x =\frac{\pi}{12}[/tex] and [tex]x = \frac{23\pi}{12}[/tex] .
