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Question Solve the trigonometric equation tan⁡θ=−3\tan \theta = -\sqrt{3} using the unit circle.

Studdy Solution

STEP 1

Assumptions
1. The unit circle is a circle with a radius of 1 centered at the origin (0,0) in the coordinate plane.
2. The tangent of an angle in the unit circle is the ratio of the y-coordinate to the x-coordinate of the point where the terminal side of the angle intersects the unit circle.
3. The equation to solve is tan⁡θ=−3\tan \theta = -\sqrt{3}.
4. The solutions for θ\theta must be within the interval [0,2π)[0, 2\pi) or [0,360∘)[0, 360^\circ) since we are dealing with the unit circle.

STEP 2

Understand the properties of the tangent function on the unit circle.
The tangent function is positive in the first and third quadrants, where the signs of the sine and cosine functions (y-coordinate and x-coordinate, respectively) are the same. It is negative in the second and fourth quadrants, where the signs of sine and cosine are different.

STEP 3

Identify the reference angle for tan⁡θ=3\tan \theta = \sqrt{3}.
The reference angle for tan⁡θ=3\tan \theta = \sqrt{3} is 60∘60^\circ or π3\frac{\pi}{3} because tan⁡60∘=tan⁡π3=3\tan 60^\circ = \tan \frac{\pi}{3} = \sqrt{3}.

STEP 4

Determine the quadrants where tan⁡θ\tan \theta is negative.
Since tan⁡θ\tan \theta is negative, the solutions must be in the second or fourth quadrants.

STEP 5

Find the angles in the second and fourth quadrants that have a reference angle of 60∘60^\circ or π3\frac{\pi}{3}.
In the second quadrant, the angle is 180∘−60∘=120∘180^\circ - 60^\circ = 120^\circ or π−π3=2π3\pi - \frac{\pi}{3} = \frac{2\pi}{3}.
In the fourth quadrant, the angle is 360∘−60∘=300∘360^\circ - 60^\circ = 300^\circ or 2π−π3=5π32\pi - \frac{\pi}{3} = \frac{5\pi}{3}.

STEP 6

Convert the angles to radians if necessary.
Since we are using the unit circle, it is common to express angles in radians.
The angle in the second quadrant is 2π3\frac{2\pi}{3} radians.
The angle in the fourth quadrant is 5π3\frac{5\pi}{3} radians.

STEP 7

Write the final solutions for θ\theta.
The solutions for the equation tan⁡θ=−3\tan \theta = -\sqrt{3} are θ=2π3\theta = \frac{2\pi}{3} and θ=5π3\theta = \frac{5\pi}{3}.

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