Respuesta :

[tex]\\ \rm\rightarrowtail 2-sin^2x=2cos^2\left(\dfrac{x}{2}\right)[/tex]

[tex]\\ \rm\rightarrowtail 1+1-sin^2x=2cos^2\left(\dfrac{x}{2}\right)[/tex]

  • 1-sin^2x=cos^2x

[tex]\\ \rm\rightarrowtail 1+cos^2x=2cos^2\left(\dfrac{x}{2}\right)[/tex]

[tex]\\ \rm\rightarrowtail cos^2x=2cos^2\left(\dfrac{x}{2}\right)-1[/tex]

[tex]\\ \rm\rightarrowtail cos^2x=cosx[/tex]

[tex]\\ \rm\rightarrowtail cosx=1[/tex]

[tex]\\ \rm\rightarrowtail x=n\pi[/tex]

As x lies in between 0 and 2π

  • The answers are 0,2π
semsee45

Answer:

[tex]x=0,\dfrac{\pi}{2}, \dfrac{3\pi}{2},2\pi[/tex]

Step-by-step explanation:

Given equation:

[tex]2-\sin^2x=2 \cos^2(\frac{x}{2})[/tex]

Subtract [tex]2 \cos^2(\frac{x}{2})[/tex] from both sides:

[tex]\implies 2-\sin^2x-2 \cos^2(\frac{x}{2})=0[/tex]

Pythagorean identity

[tex]\sin^2x+\cos^2x =1[/tex]  

[tex]\implies \sin^2x =1-\cos^2x[/tex]

Substituting into the equation:

[tex]\implies 2-(1-\cos^2x)-2 \cos^2(\frac{x}{2})=0[/tex]

[tex]\implies 1+\cos^2x-2 \cos^2(\frac{x}{2})=0[/tex]

Cosine double angle identity:

[tex]\cos(2x)=2 \cos^2x-1[/tex]

[tex]\implies 2 \cos^2x=\cos(2x)+1[/tex]

Substituting [tex]\frac{x}{2}[/tex] for [tex]x[/tex]:

[tex]\implies 2 \cos^2(\frac{x}{2})=\cos(2\cdot\frac{x}{2})+1[/tex]

[tex]\implies 2 \cos^2(\frac{x}{2})=\cos x+1[/tex]

Substituting into the equation:

[tex]\implies 1+\cos^2x- (\cos x +1)=0[/tex]

[tex]\implies \cos^2x- \cos x =0[/tex]

Factoring:

[tex]\implies \cos x(\cos x- 1)=0[/tex]

Therefore,

[tex]\cos x=0[/tex]  

[tex]\implies x=\dfrac{\pi}{2}, \dfrac{3\pi}{2}[/tex]

and  [tex]\cos x-1=0[/tex]

[tex]\implies \cos x=1[/tex]

[tex]\implies x=0, 2\pi[/tex]

Otras preguntas