Question

Solve the equation for 0 if 0° < < 360°. 2 cos 0+1= sec 0

Answer 1

Answer: To solve the equation 2cos(θ) + 1 = sec(θ), where 0° < θ < 360°, we can start by manipulating the equation using trigonometric identities.

First, we need to express sec(θ) in terms of cos(θ):

• sec(θ) = 1/cos(θ)

Now, substitute this expression back into the equation:

• 2cos(θ) + 1 = 1/cos(θ)

To eliminate the fraction, we can multiply both sides of the equation by cos(θ):

• 2cos^2(θ) + cos(θ) = 1

Now, rearrange the equation to form a quadratic equation:

• 2cos^2(θ) + cos(θ) - 1 = 0

To solve this quadratic equation, let's substitute cos(θ) with a variable, let's say, x:

• 2x^2 + x - 1 = 0

Now we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. In this case, we'll use the quadratic formula:

• x = (-b ± √(b^2 - 4ac)) / (2a)

For the equation 2x^2 + x - 1 = 0, the values of a, b, and c are:

• a = 2
• b = 1
• c = -1

Substituting these values into the quadratic formula:

• x = (-1 ± √(1^2 - 4 * 2 * -1)) / (2 * 2)

Simplifying further:

• x = (-1 ± √(1 + 8)) / 4
• x = (-1 ± √9) / 4
• x = (-1 ± 3) / 4

This gives us two possible solutions for x:

• x = (-1 + 3) / 4 = 2 / 4 = 1/2
• x = (-1 - 3) / 4 = -4 / 4 = -1

Since we are looking for values of cos(θ), we can substitute x back into cos(θ):

• cos(θ) = 1/2
• cos(θ) = -1

Now, we need to find the corresponding values of θ within the given range of 0° < θ < 360°.

For cos(θ) = 1/2, θ can be either 60° or 300° (since cos(60°) = cos(300°) = 1/2).

For cos(θ) = -1, θ can be either 180° or 360° (since cos(180°) = cos(360°) = -1).

Therefore, the solutions for the equation 2cos(θ) + 1 = sec(θ) in the given range are:

θ = 60°, 180°, 300°, 360°.