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Use an identity to solve the equation on the interval [0,2π). What is the solution to the interval?

Science & Mathematics by Anonymous 2018-07-29 13:08:13

Social Science

Use an identity to solve the equation on the interval [0,2π). What is the solution to the interval?

3 answers

4cos^2 x = 2 + 2sin x

  • Anonymous

    let w = sin x....then you have 4 - 4 w² = 2 + 2 w ===> w = 1 / 2 or -1 ===> x = π / 6 , 5π / 6 , π

  • Anonymous

    cos²(x) = 1-sin²(x) 4cos²(x) = 2 + 2sin(x) 4(1-sin²(x)) = 2 + 2sin(x) 4-4sin²(x) = 2 + 2sin(x) 4sin²(x) + 2sin(x) - 2 = 0 Quadratic formula sin(x) = [-2 ± √(2² – 4·4(-2))] / [2·4]  = [-2 ± √36] / 8  = [-2 ± 6] /8  = -1, 0.5 x = arcsin(-1), arcsin(0.5) = 3π/2, π/6, 5π/6

  • Anonymous

    0 = -2 + 2sin x +4sin^2(x) (4sinx-2)(sinx+1) = 0 sinθ θ = -1 or 1/2 θ = π/6, 5π/6, 3π/2

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