Solving Trigonometric Equations Involving a Multiple of an Angle An equation is given. (a) Find

Solving Trigonometric Equations Involving a Multiple of an...

Key Concepts

Trigonometric Equations

Trigonometric equations are equations that involve trigonometric functions like sine, cosine, and tangent. Solving these equations often requires manipulating and simplifying expressions using trigonometric identities, and finding all angles that satisfy the equation. The process frequently involves isolating a trigonometric function and then applying inverse functions or known identities to determine solutions.

Multiple-Angle Equations

Multiple-angle equations involve trigonometric functions whose arguments are multiples of the variable, such as 2? in this case. These equations require special consideration because the multiplication factors change the period of the trigonometric functions. Solving multiple-angle equations involves first addressing the equation with respect to the multiple of the variable, and then adjusting the resulting solutions to obtain the correct answers for the original variable.

Tangent Function

The tangent function is defined as the ratio of the sine of an angle to the cosine of the angle. When an equation relates sine and cosine directly, such as sin(2?)=3cos(2?), dividing both sides by cosine (provided cosine is not zero) converts the equation to an equation involving the tangent function. Recognizing this helps simplify the process by reducing the equation to a familiar form, tan(2?)=3.

General Solution for Tangent Equations

Once an equation has been simplified to the form tan(x)=constant, the general solution can be expressed as x = arctan(constant) + n?, where n is an integer. This accounts for the periodicity of the tangent function, which has a period of ?. Utilizing this general solution provides a formula for finding all possible solutions of the original equation.

Interval Restrictions

After determining the general solution for a trigonometric equation, it is often necessary to identify which solutions fall within a specified interval—commonly [0, 2?) for one complete cycle. This requires substituting integer values into the general solution to determine which solutions satisfy the interval criteria, ensuring that only the valid answers within the given domain are considered.

Transcript

00:01 All right, we're solving sine of 2 theta equals 3 times the cosine of 2 theta.

00:06 And to do this one, i notice they both have 2 theta in their arguments.

00:10 I'm just going to simplify the problem a little bit by replacing 2 theta with x, just temporarily to help me through some of the solving.

00:19 The problem now looks like sine of x equals 3 times cosine of x.

00:25 All right, so it's difficult for me to turn a sign into a cosine or vice versa with identities unless they're squared.

00:36 So the next thing i'd like to try is squaring both sides.

00:43 And we don't want to forget to square that three when we apply the square to both sides.

00:48 By turning this into a square, i can now use an identity for sine squared and replace it with 1 minus cosine squared of x.

01:00 This allows me to have all of the same type of trig function.

01:05 I can add this cosine squared of x to both sides, and that leaves me with 1 equals 10 cosine squared of x.

01:21 Okay, next i'm going to continue this solving, it's basic solving from here, dividing both sides by 10, and then last thing is the cosine of x will equal positive, or negative square root of one tenth or one over the square root of 10.

01:44 Okay, so now in order to actually get the x values that will make this work, i'm going to need a calculator.

01:51 And i like to estimate my answers first by thinking where will cosine be positive 1 over root 10? and i don't know the exact locations, i'm just estimating.

02:02 And then i would find out where cosine would be negative 1 over root 10.

02:07 And so i'm going to have answers in each of the four quadrants.

02:10 And to get the very first answer, i'm going to use a reference angle by typing in the inverse cosine of 1 over the square root of 10.

02:24 And when you enter that in your calculator, you get 1 .249.

02:33 Now i can identify what the other angles are from there.

02:36 So that gave me that first quadrant angle.

02:41 If i want to get the second quadrant angle, it would be found by doing pi and subtracting the reference angle that i just found.

02:51 And if i type that calculation in my calculator, then i get 1 .893.

03:00 Now, i've totally ignored the two theta that the original problem started with...

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