/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 Use an identity to solve each eq... [FREE SOLUTION] | 91Ó°ÊÓ

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Use an identity to solve each equation on the interval \([0,2 \pi)\) $$\sin 2 x=\cos x$$

Short Answer

Expert verified
The solutions for \(x\) in the given range \([0,2\pi)\) of the equation \(\sin 2x=\cos x\) are \(x=\frac{\pi}{6}\), \(x=\frac{5 \pi}{6}\).

Step by step solution

01

Apply trigonometric identity

The first step is to recognize which trigonometric identity can be applied to \(\sin 2x\). We know that \(\sin 2x = 2 \sin x \cos x\), so use this identity to rewrite the equation as \(2 \sin x \cos x=\cos x\).
02

Solve for \(\sin x\)

The next step is to isolate \(\sin x\) on one side of the equation. Do this by dividing both sides by \(2 \cos x\), resulting in \(\sin x=\frac{1}{2}\). Here, keep in mind that \(\cos x\) cannot equal zero because you cannot divide by zero.
03

Find possible values of \(x\)

Now using the unit circle or the chart of sine values between \(0\) and \(2\pi\), we can see that \(\sin x=\frac{1}{2}\) when \(x=\frac{\pi}{6}, \frac{5\pi}{6}\). These are the possible values for \(x\) in solution.
04

Validate with original equation

It is important to cross-check if these values satisfy the original equation \(\sin 2x=\cos x\). When \(x=\frac{\pi}{6}\), both sides equal \(\frac{\sqrt{3}}{2}\). When \(x=\frac{5\pi}{6}\), both sides equal to \(-\frac{\sqrt{3}}{2}\). So, both solutions are valid in the given range.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Identities
Understanding trigonometric identities is crucial to solving complex trigonometric equations efficiently. These identities are equations that relate the trigonometric functions to one another and can simplify expressions or make them more manageable for solving. A fundamental identity used in solving the given exercise is the double angle identity for sine, which states that \( \sin 2x = 2 \sin x \cos x \).

Applying this identity correctly transforms the original equation into a more familiar form, enabling students to isolate and solve for the sine or cosine of a single angle, which significantly simplifies the process. It's also important to remember other key trigonometric identities, such as the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \) and angle sum identities like \( \sin(x+y) = \sin x \cos y + \cos x \sin y \), as these are often utilized in solving trigonometric equations.
Unit Circle
The unit circle is an essential tool for understanding trigonometry and supporting the visualization of sine and cosine values. It's a circle with a radius of one unit with its center at the origin of a coordinate system. Angles are measured from the positive x-axis (0 radians) in a counter-clockwise direction and can use radians or degrees.

The unit circle defines the relationship between angles and the lengths of sides in a right triangle. Specifically, for any angle \( \theta \), the coordinates \( (x, y) \) of the point where the angle's terminal side intersects the circle give the cosine and sine values, respectively (\( \cos \theta \), \( \sin \theta \)). In the exercise solution, the unit circle aids in determining that the angle values for \( \sin x = \frac{1}{2} \) fall at \( \frac{\pi}{6} \) and \( \frac{5\pi}{6} \) within the given range of \( [0, 2\pi) \) because these are the points on the unit circle where the y-coordinate has a value of \( \frac{1}{2} \) and x is positive for the first value, and negative for the second, reflecting their respective quadrants.
Sine Values
In trigonometry, the sine function relates an angle of a right-angled triangle to the ratio of the length of the opposite side to the hypotenuse. Sine values for certain key angles are critical to know and can often be understood and memorized in conjunction with the unit circle.

The exercise in question involves the sine value of \( \frac{1}{2} \), which corresponds to the angles \( \frac{\pi}{6} \) and \( \frac{5\pi}{6} \), reflecting the symmetry and periodic properties of the sine function. These particular solutions can be found by relating them to the standard positions on the unit circle. It's also vital for students to recognize that the sine function has a range from \( -1 \) to \( 1 \) and observe its periodicity – specifically, that it repeats every \( 2\pi \) radians or 360 degrees. This knowledge helps identify possible angle solutions in a given range, such as \( [0, 2\pi) \) for the current exercise.

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Most popular questions from this chapter

Suppose you are solving equations in the interval \([0,2 \pi)\) Without actually solving equations, what is the difference between the number of solutions of \(\sin x=\frac{1}{2}\) and \(\sin 2 x=\frac{1}{2} ?\) How do you account for this difference?

Find the exact value of each expression. Do not use a calculator. $$\sin \left(\cos ^{-1} \frac{1}{2}+\sin ^{-1} \frac{3}{5}\right)$$

Will help you prepare for the material covered in the next section. Use the appropriate values from Exercise 101 to answer each of the following. a. Is \(\sin \left(30^{\circ}+60^{\circ}\right),\) or \(\sin 90^{\circ},\) equal to \(\sin 30^{\circ}+\sin 60^{\circ} ?\) b. Is \(\sin \left(30^{\circ}+60^{\circ}\right),\) or \(\sin 90^{\circ},\) equal to \(\sin 30^{\circ} \cos 60^{\circ}+\cos 30^{\circ} \sin 60^{\circ} ?\)

Use this information to solve. When throwing an object, the distance achieved depends on its initial velocity, \(v_{0}\) and the angle above the horizontal at which the object is thrown, \(\theta\) The distance, \(d\), in feet, that describes the range covered is given by $$d=\frac{v_{0}^{2}}{16} \sin \theta \cos \theta$$ where \(v_{0}\) is measured in feet per second. You and your friend are throwing a baseball back and forth. If you throw the ball with an initial velocity of \(v_{0}=90\) feet per second, at what angle of elevation, \(\theta,\) to the nearest degree, should you direct your throw so that it can be easily caught by your friend located 170 feet away?

Use a graphing utility to approximate the solutions of each equation in the interval \([0,2 \pi) .\) Round to the nearest hundredth of a radian. $$2 \sin ^{2} x=1-2 \sin x$$

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