Question

solve on the interval [0, 2pi). 1.2sin^2(x)+cos(x)=1

Answer 1

Question #9992

Solve on the interval [0, 2pi). 1.2sin^2(x)+cos(x)=1. Please show your work/

Solution

1,2sin²x + cos x = 1;

1,2(1 - cos²x) + cos x - 1 = 0;

1,2 - 1,2cos²x + cos x - 1 = 0;

1,2cos²x - cos x - 0,2 = 0;

cos x = t;

t ∈ [-1;1]

1,2t² - t - 0,2 = 0;

√D = 1,4;

t₁ = 1; t₂ = -{1/6};

{

cos x = 1

}

{

cos x = -{1/6

}

x₁ = 0;

x₂ = arccos(-{1/6});

x₃ = -arccos(-{1/6}).

x₁ = 0;

Answer: x₂ = arccos(-{1/6});

x₃ = -arccos(-{1/6}).