Mathematics
gloriallanas
11

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

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(2) Answers
Gakley64

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=+-[latex] \frac{ \pi }{6}+2 \pi k [/latex], k∈Z find value for the interval [0, 2 pi ) {[latex] \frac{ \pi }{6}, \pi , - \frac{ \pi }{6}+2 \pi k [/latex]} Answer: [latex] \frac{ \pi }{6}, \pi , \frac{ 11\pi }{6}[/latex]

indiapowell86

[latex]2cos^{2}(x) + cos(x)-1 = 0[/latex] This could also be written as, where [latex]a = cos(x)[/latex] [latex]2a^{2} + a - 1 = 0[/latex] This would factorize to give:  [latex](2a-1)(a+1)=0[/latex] So we can factorize our original expression: [latex]2cos^{2}(x) + cos(x)-1 = 0 \\ \\ (2cosx - 1)(cosx+1) = 0 [/latex] We can then solve for [latex]x[/latex] as we would with a normal quadratic: [latex]2cosx -1 =0 \\ \\ cosx = \frac{1}{2} \\ \\ x = cos^{-1}( \frac{1}{2} ) \\ \\ x = \frac{ \pi }{3}, \frac{5 \pi }{3} [/latex] And also: [latex]cos(x)+1 = 0 \\ \\ cos(x)= -1 \\ \\ x = cos^{-1}(1) \\ \\ x = 0, 2 \pi [/latex] So our values for [latex]x[/latex] are: [latex]x =0, \frac{ \pi }{3}, \frac{5 \pi }{3}, 2 \pi[/latex] As: [latex]0 \leq x\ \textless \ 2 \pi [/latex] Our final solutions for [latex]x[/latex] are: [latex]x = \boxed{0, \frac{ \pi }{3}, \frac{5 \pi }{3}}[/latex]

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