Solve the given trigonometric equation on 0^∘ ≤θ<360^∘ and express the answer in degrees rounded

Solve the given trigonometric equation on 0^∘ ≤θ<360^∘ and...

Key Concepts

Quadratic Equations in Trigonometric Functions

This concept involves recognizing and rewriting a trigonometric equation as a quadratic by using a substitution (for example, letting u = sin(x) or a similar function). It allows one to apply familiar techniques from algebra, such as factoring or the quadratic formula, to solve for the trigonometric expression, which is then substituted back in to find the variable of interest.

Substitution Method

The substitution method is an approach used to simplify trigonometric equations by replacing a repeated trigonometric expression with a single variable. This reduces the complexity of the equation, making it easier to solve. After solving the simplified equation, the substitution is reversed to obtain the solution in terms of the original variable.

Inverse Trigonometric Functions

Inverse trigonometric functions are used to find angle measures when the value of a trigonometric function is known. After isolating the trigonometric expression, applying the appropriate inverse function gives the principal solution, which must then be augmented by the function's periodicity to obtain all solutions.

Periodicity of Trigonometric Functions

Trigonometric functions are periodic, meaning that they repeat their values in regular intervals. When solving trigonometric equations, it is essential to account for this periodicity by finding all angles within the given domain that satisfy the equation. This might involve adding integer multiples of the period to the principal solution.

Numerical Rounding

Numerical rounding is the process of adjusting numbers to a specified degree of precision. In many trigonometric problems, especially those requiring a numeric answer, the final results need to be rounded to a desired decimal place. This ensures consistency and clarity in the presentation of the solutions.

Transcript

00:01 All right, so this problem, we're looking at 15 sine squared of 2 theta plus sign of 2 theta minus 2 equals 0.

00:20 So i'm looking at this 15 minus 2.

00:22 If i multiply those together, it gives me negative 30.

00:25 So factors of that to give me 1 is 6 minus 5.

00:31 And that's kind of my clue as to how to get these answers.

00:35 Because there's only one way to get two, and that's two times one.

00:38 And then two times three gives me six.

00:42 I might throw on a sign in here of two theta.

00:46 And that term needed to be positive.

00:49 And then one times negative one times five sign of two theta.

00:57 And you can double check this as correct, because if you foil this out, you do get this top piece.

01:05 So we're good.

01:06 As far.

01:07 And if you're second -guessing my method, there's so many other ways of factoring, like, understanding how to factor this that i don't want to get into it.

01:15 That would take me a half hour to teach us so many different ways.

01:21 Anyway, so this answer now, set this equal to zero, subtract two and divide by five, and then this other answer, sign of two theta, set it equal to zero, add one and divide by three.

01:34 Now, this is where it gets complicated.

01:36 Because now to solve each of these problems, you have to do the inverse sign and make sure you're in degree mode.

01:45 So second sign and double check you're in the correct mode in your calculator.

01:49 So as i do this, inverse sign of negative two -fiths, i get an answer of negative 23 .58 degrees.

02:01 Definitely wrong because we only want positive answers.

02:04 So how do we fix that? we add 360 degrees to that answer because your calculator only goes clockwise here.

02:12 So add 360 to that answer.

02:17 So it could be 336 .42.

02:22 Okay.

02:23 Add another 360 to it.

02:25 And you might be sitting there and saying, why are you adding another 360? that's too big.

02:29 I still have to divide by two here.

02:32 And why this is even more complicated is sine could have been negative in this quadrant as well.

02:39 So i need to do 180 plus that 23 .58 to see that that angle could have also been 203 .58.

02:50 Or adding 360 to that answer could have been 563 .58.

02:58 And again, why do we have to do all this other stuff? we have to divide all those numbers by two just to figure out that one answer.

03:06 So theta could be 336...

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