Find all solutions to the equation. tanθ=1\tan\theta=1tanθ=1

θ = π 4 + π n \theta=\frac{\pi}{4}+\pi n θ = 4 π + πn

Find all solutions to the equation. tan θ = 1 \tan\theta=1 tan θ = 1

this problem, we're asked to find all of the solutions to the equation, the tangent of theta equals 1. So the first thing we want to do is look at our unit circle and find all of the solutions to this equation that are on our unit circle. So looking at that unit circle, I know that for pi over 4, my tangent is equal to 1. And then also in quadrant 3, that's the other quadrant for which tangent values are positive, I know that for 5 pi over 4, my tangent is also equal to 1. So my two angles, my two solutions that are on that unit circle are pi over 4 and 5pi over 4. Now for these solutions, I can go ahead and just add 2pi n to each of them in order to represent all of the possible solutions to this equation. Now something that you might notice here is that looking at these two angles, they are exactly half of a rotation or pi radians away from each other. Now because of that, I can actually further simplify my answer. So this is technically correct, but we can get this even more concise instead of having these 2 long expressions as our solutions. So because these angles are pi radians apart, I can actually simplify this to pi over 4 plus pinn. Then if I start with my angle, pi over 4, and I add a pi, and then I add another pi, and another pi, and another pi, I will keep going around to each of these two angles that I've identified on my unit circle to get an infinite number of solutions at these two locations. So I don't have to worry about having these 2 long solutions because I can express all of my solutions in their most simplified form as beta equals pi over 4+pi n. Thanks for watching, and let me know if you have questions.

Find all solutions to the equation. tan ⁡ θ = 1 \tan\theta=1 tan θ = 1

θ = π 4 + π n \theta=\frac{\pi}{4}+\pi n θ = 4 π + πn

θ = 2 π n \theta=2\pi n θ = 2 πn

θ = π 4 + 2 π n \theta=\frac{\pi}{4}+2\pi n θ = 4 π + 2 πn

θ = 5 π 4 + 2 π n \theta=\frac{5\pi}{4}+2\pi n θ = 4 5 π + 2 πn

5. Inverse Trigonometric Functions and Basic Trigonometric Equations

Linear Trigonometric Equations

Trigonometry

5. Inverse Trigonometric Functions and Basic Trigonometric Equations

Linear Trigonometric Equations

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Multiple Choice

Find all solutions to the equation.
$\tan\theta=1$

$\theta=2\pi n$

$\theta=\frac{\pi}{4}+2\pi n$

$\theta=\frac{\pi}{4}+\pi n$

$\theta=\frac{5\pi}{4}+2\pi n$

Verified step by step guidance

Understand that the equation $ \tan\theta = 1 $ implies that $ \theta $ is an angle where the tangent function equals 1. This occurs at specific angles on the unit circle.

Recall that the tangent function has a period of $ \pi $, meaning it repeats every $ \pi $ radians. Therefore, if $ \tan\theta = 1 $ at some angle $ \theta_0 $, it will also be true at $ \theta_0 + \pi n $ for any integer $ n $.

Identify the principal angle where $ \tan\theta = 1 $. This occurs at $ \theta = \frac{\pi}{4} $ because $ \tan\left(\frac{\pi}{4}\right) = 1 $.

Since the tangent function is periodic with period $ \pi $, the general solution for $ \theta $ is $ \theta = \frac{\pi}{4} + \pi n $, where $ n $ is any integer.

Additionally, consider the angle $ \theta = \frac{5\pi}{4} $ where $ \tan\theta = 1 $ again, due to the periodicity of the tangent function. Thus, another set of solutions is $ \theta = \frac{5\pi}{4} + 2\pi n $, where $ n $ is any integer.

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