17-24 Solving Basic Trigonometric Equations Solve the given equation, and list six specific solu

Name: Problem 20, Chapter 7: Algebra and Trigonometry
Uploaded: 2019-06-23T17:15:59-08:00
Duration: 4 min 52 s
Channel: Mihir Nayar
Description: 17-24 Solving Basic Trigonometric Equations Solve the given equation, and list six specific solutions. sinθ=-(√(3))/(2)

17-24 Solving Basic Trigonometric Equations Solve the given...

Key Concepts

Unit Circle

The unit circle is a fundamental concept in trigonometry that relates angles to points on a circle with a radius of one. It helps in visualizing and determining the values of sine, cosine, and other trigonometric functions for different angles, serving as a primary tool for solving trigonometric equations.

Reference Angle

A reference angle is the acute angle formed by the terminal side of a given angle and the horizontal axis. It is used to relate an angle in any quadrant to its corresponding acute angle in the first quadrant, which is useful for finding all angles that have the same trigonometric values up to sign changes.

Periodicity

Periodicity refers to the property of trigonometric functions where they repeat their values in regular intervals. This characteristic is crucial when solving equations because it allows for the construction of general solutions by adding integer multiples of the period to any particular solution.

General Solution

The concept of a general solution in trigonometric equations involves finding not just individual angles that satisfy the equation, but a complete set of solutions expressed in a formula that encompasses all angles obtained by adding the periodic interval. This ensures that all possible answers are accounted for.

Transcript

00:01 Hey guys, today we're solving problem number 20 on page 605 with the textbook, which asks us, excuse me, to find all solutions for theta in the trigonometric equation, the sign of theta equals negative root 3 over 2.

00:15 And it also asks us to list six specific solutions.

00:21 So the first thing you're going to want to do is to notice that you're dealing with the root 3 over 2 and a negative sign.

00:27 So you're working with quadrants three and four and a sign sign.

00:32 So sign is negative in quadrants three and four, which is why you're only dealing with those two quadrants.

00:42 So we have a negative sign and we have a root three over two.

00:46 So you're literally, if you've memorized unit circle, you would already know that at 4 pi over three, you would have a negative root three over two for sign data.

00:54 And then same thing at 5 pi over three.

01:00 So general solutions, pretty easy.

01:04 And also remember at 4 pi over 3, when you go back around 2 pi radians to 10 pi over 3, you would end up at that same spot with that sign of theta being negative root 3 over 2, and at 5 pi over 3, if we went back around 6 pi over 3 or 2, again, you'd end up at 11 pi over 3, but you'd still have that sign of theta equal to negative root 3 over 2.

01:26 So theta, generally for 4 pi over 3, it's going to be 4 pi over 3 plus 2 pi.

01:39 Again, that 2 pi comes from our 2 pi radians or 360 degrees, which is one full circle rotation.

01:46 So 2 pi times k, where k is some sort of integer value, and then at 5 pi over 3, you would have theta equals 5 pi over 3 plus that same 2 pi.

02:04 Again because that's a full circle rotation, which is what we want to get back to that same sign of theta that would yield a negative root 3 over 2.

02:18 So again, we would have times k.

02:23 So there are your two general solutions.

02:25 And then they asks us to list six specific solutions.

02:29 So again, k is any integer value.

02:31 So i'm just going to let k be 0, 1, and 2.

02:34 So at 0, you would have, so 6.

02:38 Specific solutions, theta, at zero for 4 pi every 3, you simply have 4 pi over 3...

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