Solved on Dec 14, 2023

Solve the trigonometric equation $\sec^2 \theta + 2 \sec \theta = 0$ for $\theta$ .

STEP 1

Assumptions
1. We are solving the equation $\sec^2 \theta + 2 \sec \theta = 0$ for $\theta$ .
2. $\sec \theta$ is the secant function, which is the reciprocal of the cosine function, so $\sec \theta = \frac{1}{\cos \theta}$ .
3. The domain of $\sec \theta$ excludes values of $\theta$ for which $\cos \theta = 0$ , since division by zero is undefined.

STEP 2

Let's introduce a substitution to simplify the equation. Let $x = \sec \theta$ , which means $x = \frac{1}{\cos \theta}$ .

STEP 3

Rewrite the original equation using the substitution $x = \sec \theta$ .
$x^2 + 2x = 0$

STEP 4

Factor out $x$ from the equation.
$x(x + 2) = 0$

STEP 5

Apply the zero product property, which states that if a product of factors is zero, then at least one of the factors must be zero.
$x = 0 \quad \text{or} \quad x + 2 = 0$

STEP 6

Solve the first equation $x = 0$ . However, since $x = \sec \theta$ and $\sec \theta$ cannot be zero (because $\cos \theta$ cannot be infinite), this solution is not valid within the domain of $\sec \theta$ .

STEP 7

Solve the second equation $x + 2 = 0$ .
$x = -2$

STEP 8

Substitute back $\sec \theta$ for $x$ .
$\sec \theta = -2$

STEP 9

Recall that $\sec \theta = \frac{1}{\cos \theta}$ , so we can write:
$\frac{1}{\cos \theta} = -2$

STEP 10

Take the reciprocal of both sides to solve for $\cos \theta$ .
$\cos \theta = -\frac{1}{2}$

STEP 11

Determine the values of $\theta$ for which $\cos \theta = -\frac{1}{2}$ . These values are found by considering the unit circle and the symmetry of the cosine function.
$\theta = \frac{2\pi}{3} + 2k\pi \quad \text{or} \quad \theta = \frac{4\pi}{3} + 2k\pi, \quad k \in \mathbb{Z}$
The solutions for $\theta$ are $\frac{2\pi}{3} + 2k\pi$ and $\frac{4\pi}{3} + 2k\pi$ , where $k$ is any integer.

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Solve the trigonometric equation $\sec^2 \theta + 2 \sec \theta = 0$ for $\theta$ .

STEP 1

STEP 2

Let's introduce a substitution to simplify the equation. Let $x = \sec \theta$ , which means $x = \frac{1}{\cos \theta}$ .

STEP 3

Rewrite the original equation using the substitution $x = \sec \theta$ .
$x^2 + 2x = 0$

STEP 4

Factor out $x$ from the equation.
$x(x + 2) = 0$

STEP 5

Apply the zero product property, which states that if a product of factors is zero, then at least one of the factors must be zero.
$x = 0 \quad \text{or} \quad x + 2 = 0$

STEP 6

Solve the first equation $x = 0$ . However, since $x = \sec \theta$ and $\sec \theta$ cannot be zero (because $\cos \theta$ cannot be infinite), this solution is not valid within the domain of $\sec \theta$ .

STEP 7

Solve the second equation $x + 2 = 0$ .
$x = -2$

STEP 8

Substitute back $\sec \theta$ for $x$ .
$\sec \theta = -2$

STEP 9

Recall that $\sec \theta = \frac{1}{\cos \theta}$ , so we can write:
$\frac{1}{\cos \theta} = -2$

STEP 10

Take the reciprocal of both sides to solve for $\cos \theta$ .
$\cos \theta = -\frac{1}{2}$

STEP 11

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Solve the trigonometric equation sec⁡2θ+2sec⁡θ=0\sec^2 \theta + 2 \sec \theta = 0sec2θ+2secθ=0 for θ\thetaθ.

STEP 1

STEP 2

Let's introduce a substitution to simplify the equation. Let x=sec⁡θx = \sec \thetax=secθ, which means x=1cos⁡θx = \frac{1}{\cos \theta}x=cosθ1​.

STEP 3

Rewrite the original equation using the substitution x=sec⁡θx = \sec \thetax=secθ.x2+2x=0x^2 + 2x = 0x2+2x=0

STEP 4

Factor out xxx from the equation.x(x+2)=0x(x + 2) = 0x(x+2)=0

STEP 5

Apply the zero product property, which states that if a product of factors is zero, then at least one of the factors must be zero.x=0orx+2=0x = 0 \quad \text{or} \quad x + 2 = 0x=0orx+2=0

STEP 6

Solve the first equation x=0x = 0x=0. However, since x=sec⁡θx = \sec \thetax=secθ and sec⁡θ\sec \thetasecθ cannot be zero (because cos⁡θ\cos \thetacosθ cannot be infinite), this solution is not valid within the domain of sec⁡θ\sec \thetasecθ.

STEP 7

Solve the second equation x+2=0x + 2 = 0x+2=0.x=−2x = -2x=−2

STEP 8

Substitute back sec⁡θ\sec \thetasecθ for xxx.sec⁡θ=−2\sec \theta = -2secθ=−2

STEP 9

Recall that sec⁡θ=1cos⁡θ\sec \theta = \frac{1}{\cos \theta}secθ=cosθ1​, so we can write:1cos⁡θ=−2\frac{1}{\cos \theta} = -2cosθ1​=−2

STEP 10

Take the reciprocal of both sides to solve for cos⁡θ\cos \thetacosθ.cos⁡θ=−12\cos \theta = -\frac{1}{2}cosθ=−21​

STEP 11

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Solve the trigonometric equation $\sec^2 \theta + 2 \sec \theta = 0$ for $\theta$ .

Let's introduce a substitution to simplify the equation. Let $x = \sec \theta$ , which means $x = \frac{1}{\cos \theta}$ .

Rewrite the original equation using the substitution $x = \sec \theta$ .
$x^2 + 2x = 0$

Factor out $x$ from the equation.
$x(x + 2) = 0$

Apply the zero product property, which states that if a product of factors is zero, then at least one of the factors must be zero.
$x = 0 \quad \text{or} \quad x + 2 = 0$

Solve the first equation $x = 0$ . However, since $x = \sec \theta$ and $\sec \theta$ cannot be zero (because $\cos \theta$ cannot be infinite), this solution is not valid within the domain of $\sec \theta$ .

Solve the second equation $x + 2 = 0$ .
$x = -2$

Substitute back $\sec \theta$ for $x$ .
$\sec \theta = -2$

Recall that $\sec \theta = \frac{1}{\cos \theta}$ , so we can write:
$\frac{1}{\cos \theta} = -2$

Take the reciprocal of both sides to solve for $\cos \theta$ .
$\cos \theta = -\frac{1}{2}$