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Solve, in degrees to 1 decimal place, for 0 ⩽ θ < 180

2 cos(2θ + 30)◦ + tan(2θ + 30)◦ = 0

User Tkr
by
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1 Answer

4 votes

Answer:

θ = 94.1 , 145.9 degrees.

Explanation:

2 cos(2θ + 30) = - tan(2θ + 30)

Divide both sides by tan(2θ + 30)

2 cos(2θ + 30)cot(2θ + 30) = -1

cos(2θ + 30)cot(2θ + 30) = -1/2

cos(2θ + 30)cot(2θ + 30) + 1/2 = 0

2(cos(2θ + 30)cot(2θ + 30) + 1) = 0

cos(2θ + 30)cot(2θ + 30) + 1 = 0

Let 2θ + 30 = X, then:

cos X cot X + 1 = 0

cos X * cos X / sin x + 1 = 0

cos^2 X / sin X + 1 = 0

(1 - sin^2 X) sin X + 1 = 0

1 - sin^2x + sin X = 0

sin^2 X - sin X - 1 = 0

sin X = 1.618, -0.618 (using the quadratic formula)

This gives X = 218.17, 321.83 degrees

So

2θ + 30 = 218.17

2θ = 188.17

θ = 94.085

and

2θ + 30 = 321.83

2θ = 291.83

θ = 145.915

User Blake Watters
by
7.3k points