Problem 46 - Contemporary Precalculus

Question

Use an appropriate substitution (as in Example 7 ) to find all solutions of the equation. $$2 \sin \frac{x}{3}=1$$

Step-by-Step Solution

Answer

Question: Find all the solutions for the equation $$2 \sin (\frac{x}{3})=1$$. Answer: All the solutions for the given equation are $$x = \frac{\pi}{2} + 6\pi n$$ or $$x = 3\pi - \frac{\pi}{2} + 6\pi n$$, where $$n$$ is an integer.

Step 1: Choose an appropriate substitution

We can use the substitution $$u = \frac{x}{3}$$ to simplify the given equation. Note that when we find the solutions for $$u$$, we should not forget to substitute back to get the solutions for $$x$$.

Step 2: Rewrite the equation with the new variable

Substitute $$u = \frac{x}{3}$$ and rewrite the equation, we get: $$2 \sin u=1$$

Step 3: Solve the simplified equation

Divide both sides by 2: $$\sin u = \frac{1}{2}$$ Now, we should find all the angles whose sine is equal to $$\frac{1}{2}$$ in the standard domain. Recall that sin(30°) = sin(π/6) = 1/2, so we have $$u = \frac{\pi}{6} + 2\pi n$$ or $$u = \pi - \frac{\pi}{6} + 2\pi n$$ where $$n$$ is an integer.

Step 4: Substitute back to find the solutions for $$x$$

Now, we should replace $$u$$ with $$\frac{x}{3}$$ to find the solutions for $$x$$: $$\frac{x}{3} = \frac{\pi}{6} + 2\pi n$$ or $$\frac{x}{3} = \pi - \frac{\pi}{6} + 2\pi n$$ Multiply both sides by 3: $$x = \frac{\pi}{2} + 6\pi n$$ or $$x = 3\pi - \frac{\pi}{2} + 6\pi n$$ Thus, all the solutions for the given equation are: $$x = \frac{\pi}{2} + 6\pi n$$ or $$x = 3\pi - \frac{\pi}{2} + 6\pi n$$, where $$n$$ is an integer.

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique used to simplify trigonometric equations. It involves replacing a trigonometric function of an angle with a simpler variable to make the equation easier to solve. The substitution not only simplifies calculations but also provides a way to visualize the problem in a more straightforward manner.

For instance, in our given exercise, the substitution chosen is \( u = \frac{x}{3} \). This step reduces the complexity of the trigonometric equation by focusing on \( u \) rather than dealing with the fraction \( \frac{x}{3} \) within the sine function.

Sine Function

The sine function, represented by \( \sin \), is one of the fundamental trigonometric functions. It relates an angle of a right-angled triangle to the ratio of the length of the opposite side to the hypotenuse.

In a unit circle context, \( \sin(\theta) \) describes the y-coordinate of the point where the terminal side of an angle \( \theta \) intersects the unit circle. Sine values range from -1 to 1, with \( \sin(\frac{\pi}{6}) \) being one of the commonly remembered values because it equals \( \frac{1}{2} \).When solving equations involving the sine function, it's essential to remember that it is periodic with a period of \( 2\pi \), meaning the function repeats its values every \( 2\pi \) radians. In our exercise, understanding this helped identify all possible angle solutions using the fact that \( \sin(\frac{\pi}{6}) = \sin(\pi - \frac{\pi}{6}) = \frac{1}{2} \) and reasoning about the sine function's periodicity.

Angle Solutions in Trigonometry

In trigonometry, finding angle solutions typically involves determining the values of angles that satisfy a given trigonometric equation. Angles can be measured in radians or degrees, and in mathematical problems, radians are often the preferred unit.

Because trigonometric functions like the sine, cosine, and tangent are periodic, there can be an infinite number of angles that satisfy an equation. Therefore, solutions are usually expressed in terms of all the angles that are co-terminal with the initial angle solutions.In our original exercise, the solution found for \( u \) was \( u = \frac{\pi}{6} + 2\pi n \) and \( u = \pi - \frac{\pi}{6} + 2\pi n \) where \( n \) is an integer. These expressions take into account the fact that if a sine value is known at a certain angle, then it will recur at that angle plus \( 2\pi n \) for any integer \( n \).Once we obtain the solution for \( u \) based on its sine value, we substitute back to find the corresponding \( x \) values in the original equation, effectively unwinding the initial substitution and yielding the final solutions. This process demonstrates the interplay of substitution and understanding the properties of trigonometric functions to solve equations efficiently.