Solve for θ in Degrees cos(theta)^2+9cos(theta)+18=0

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Trigonometry

$\cos^{2} (θ) + 9 \cos (θ) + 18 = 0$

Factor the left side of the equation.

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Let $u = \cos (θ)$ . Substitute $u$ for all occurrences of $\cos (θ)$ .

$u^{2} + 9 u + 18 = 0$

Factor $u^{2} + 9 u + 18$ using the AC method.

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Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $18$ and whose sum is $9$ .

$3, 6$

Write the factored form using these integers.

$(u + 3) (u + 6) = 0$

Replace all occurrences of $u$ with $\cos (θ)$ .

$(\cos (θ) + 3) (\cos (θ) + 6) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$\cos (θ) + 3 = 0$

$\cos (θ) + 6 = 0$

Set $\cos (θ) + 3$ equal to $0$ and solve for $θ$ .

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Set $\cos (θ) + 3$ equal to $0$ .

$\cos (θ) + 3 = 0$

Solve $\cos (θ) + 3 = 0$ for $θ$ .

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Subtract $3$ from both sides of the equation.

$\cos (θ) = - 3$

The range of cosine is $- 1 \leq y \leq 1$ . Since $- 3$ does not fall in this range, there is no solution.

No solution

The final solution is all the values that make $(\cos (θ) + 3) (\cos (θ) + 6) = 0$ true.

No solution

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