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

Question

Could someone please help me solve the equation ( 2\cos^2x + \cos x - 1 = 0 ) for ( x ) in the interval ([0, 2\pi))? 
Here are the answer choices I am considering:
a) ( \pi ) and ( \pi/3 )
b) ( \pi, \pi/3, ) and ( 5\pi/3 )
c) ( 1 ) and ( 2\pi/3 )
d) ( 1, 2\pi/3, ) and ( 4\pi/3 )
e) ( 1, \pi/3, ) and ( 5\pi/3 )  
I would greatly appreciate quick assistance in solving this!

Mr. Louisa Denesik PhD · Accepted Answer

2cos^2(x) + cos(x) &minus; 1 = 0u = cos(x)=> 2u^2 + u &minus; 1 = 0=> 2u^2 + 2u &minus; u &minus; 1 = 0=> 2u(u + 1) &minus; 1(u + 1) = 0=> (u + 1)(2u &minus; 1) = 0=> u = -1 or u = 1/2=> cos(x) = -1 or cos(x) = 1/2=> x = 2&pi;n &plusmn; cos&#8315;&sup1;(-1) or 2&pi;n &plusmn; cos&#8315;&sup1;(1/2) where n&isin;&#8484;=> x = 2&pi;n &plusmn; &pi; or 2&pi;n &plusmn; &pi;/3 where n&isin;&#8484;over the interval [0, 2&pi;)x = &pi;/3, &pi;, 5&pi;/3answer B

Anonymous · Answer

2cos^2x + cosx &minus; 1 = 0[2cos(x) - 1][cos(x) + 1] = 02cos(x) - 1 = 0, cos(x) + 1 = 02cos(x) = 1, .....cos(x) = - 1cos(x) = 1/2, x = &pi;/3, 5&pi;/3, &pi;, answer//

Garfield Ryan · Answer

If you let u = cos(x), then u&sup2; = cos&sup2;(x) and the equation becomes:2u&sup2; + u - 1 = 0This factors into:(2u - 1)(u + 1) = 0u = 1/2 and u = -1Reverse the substitution for u:cos(x) = 1/2 and cos(x) = -1You can take it from here.

Anonymous · Answer

2*cos^2(x) + cos(x) - 1 = 0
Let u = cos(x), so we have:
2u^2 + u - 1 = 0
2u^2 + 2u - u - 1 = 0
2u(u + 1) - (u + 1) = 0
(2u - 1)(u + 1) = 0
2u - 1 = 0 or u + 1 = 0
2u = 1 or u = -1
u = 1/2 or u = -1
cos(x) = 1/2 or cos(x) = -1
x = pi/3 or x = pi or x = 5*pi/3

Anonymous

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

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