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

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a= pi and pi/3b= pi, pi/3, and 5pi/3c= 1, and 2pi/3d= 1, 2pi/3, and 4pi/3 e= 1, pi/3, and 5pi/3i would appreciate quick help

Lavinia Carroll · Accepted Answer

2cos^2(x) + cos(x) − 1 = 0u = cos(x)=> 2u^2 + u − 1 = 0=> 2u^2 + 2u − u − 1 = 0=> 2u(u + 1) − 1(u + 1) = 0=> (u + 1)(2u − 1) = 0=> u = -1 or u = 1/2 => cos(x) = -1 or cos(x) = 1/2=> x = 2πn ± cos⁻¹(-1) or 2πn ± cos⁻¹(1/2) where n∈ℤ=> x = 2πn ± π or 2πn ± π/3 where n∈ℤover the interval [0, 2π)x = π/3, π, 5π/3answer B

Lavinia Carroll · Answer

2cos^2x + cosx − 1 = 0[2cos(x) - 1][cos(x) + 1] = 02cos(x) - 1 = 0, cos(x) + 1 = 02cos(x) = 1, .....cos(x) = - 1 cos(x) = 1/2, x = π/3, 5π/3, π, answer//

Lavinia Carroll · Answer

If you let u = cos(x), then u² = cos²(x) and the equation becomes:2u² + 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.

Lavinia Carroll · Answer

2*cos^2(x) + cos(x) - 1 = 0Let u = cos(x), so we have:2u^2 + u - 1 = 02u^2 + 2u - u - 1 = 02u(u + 1) - (u + 1) = 0(2u - 1)(u + 1) = 02u - 1 = 0 or u + 1 = 02u = 1 or u = -1u = 1/2 or u = -1cos(x) = 1/2 or cos(x) = -1x = pi/3 or x = pi or x = 5*pi/3

Alize Kuhn

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

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