JessicaRufusidn
Answered

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Solve 2cos^2x + cosx − 1 = 0 for x over the interval [0, 2 pi ).

Sagot :

Belcatyaidn
2cos²x + cosx − 1 = 0
(cosx+1)(2cosx-1)=0
so cosx+1=0
or 2cosx-1=0
1) cosx+1=0
cosx=-1
x=π+2πk, k∈Z

2) 2cosx-1=0
2cosx=1
cosx=0.5
x=+-[tex] \frac{ \pi }{6}+2 \pi k [/tex], k∈Z


find value for the interval [0, 2 pi )

{[tex] \frac{ \pi }{6}, \pi , - \frac{ \pi }{6}+2 \pi k [/tex]}

Answer: [tex] \frac{ \pi }{6}, \pi , \frac{ 11\pi }{6}[/tex]



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

This could also be written as, where [tex]a = cos(x)[/tex]

[tex]2a^{2} + a - 1 = 0[/tex]

This would factorize to give:

 [tex](2a-1)(a+1)=0[/tex]

So we can factorize our original expression:

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

We can then solve for [tex]x[/tex] as we would with a normal quadratic:

[tex]2cosx -1 =0 \\ \\ cosx = \frac{1}{2} \\ \\ x = cos^{-1}( \frac{1}{2} ) \\ \\ x = \frac{ \pi }{3}, \frac{5 \pi }{3} [/tex]

And also:

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

So our values for [tex]x[/tex] are:

[tex]x =0, \frac{ \pi }{3}, \frac{5 \pi }{3}, 2 \pi[/tex]

As: [tex]0 \leq x\ \textless \ 2 \pi [/tex]

Our final solutions for [tex]x[/tex] are:

[tex]x = \boxed{0, \frac{ \pi }{3}, \frac{5 \pi }{3}}[/tex]


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