Respuesta :
Solution to equation[tex]cosxtanx - 2 cos^2 x=-1[/tex] for all real values of x is [tex]x=2k\pi + \frac{\pi}{6} , x=2k\pi + \frac{5\pi}{6}[/tex] .
Step-by-step explanation:
Here we have , [tex]cosxtanx - 2 cos^2 x=-1[/tex]. Let's solve :
⇒ [tex]cosxtanx - 2 cos^2 x=-1[/tex]
⇒ [tex]cosx(\frac{sinx}{cosx}) - 2 cos^2 x=-1[/tex]
⇒ [tex]sinx = 2 cos^2 x-1[/tex]
⇒ [tex]sinx = 2 (1-sin^2x)-1[/tex]
⇒ [tex]sinx = 1-2sin^2x[/tex]
⇒ [tex]2sin^2x+sinx-1=0[/tex]
By quadratic formula :
⇒ [tex]sinx = \frac{-b \pm \sqrt{b^2-4ac} }{2a}[/tex]
⇒ [tex]sinx = \frac{-1 \pm \sqrt{1^2-4(2)(-1)} }{2(2)}[/tex]
⇒ [tex]sinx = \frac{-1 \pm3}{4}[/tex]
⇒ [tex]sinx = \frac{1}{2} , sinx =-1[/tex]
⇒ [tex]sinx = sin\frac{\pi}{6} , sinx = sin\frac{3\pi}{2}[/tex]
⇒ [tex]x=\frac{\pi}{6} , x=\frac{3\pi}{2}[/tex]
But at [tex]x=\frac{3\pi}{2}[/tex] we have equation undefined as [tex]cos\frac{3\pi}{2}=0[/tex] . Hence only solution is :
⇒ [tex]x=\frac{\pi}{6}[/tex]
Since , [tex]sin(\pi -x)=sinx[/tex]
⇒ [tex]x=\pi -\frac{\pi}{6} = \frac{5\pi}{6}[/tex]
Now , General Solution is given by :
⇒ [tex]x=2k\pi + \frac{\pi}{6} , x=2k\pi + \frac{5\pi}{6}[/tex]
Therefore , Solution to equation[tex]cosxtanx - 2 cos^2 x=-1[/tex] for all real values of x is [tex]x=2k\pi + \frac{\pi}{6} , x=2k\pi + \frac{5\pi}{6}[/tex] .
