Question

Solve the equation: \(4 \sin \theta \cos \theta=1,0 \leq \theta<2 \pi\).

Short answer

The solutions are \(\theta = \frac{\text{π}}{12}, \frac{5\text{π}}{12}, \frac{7\text{π}}{12}, \frac{11\text{π}}{12} \).

Step-by-step solution

Use Double Angle Identity

Recall that the double angle identity for sine is given by: \[ \text{sin}(2\theta) = 2 \text{sin}(\theta) \text{cos}(\theta) \]. To use this identity, observe that the equation can be written as: \[ 2 (2 \text{sin}(\theta) \text{cos}(\theta)) = 1 \]. Hence, we have: \[ 2 \text{sin}(2\theta) = 1 \].

Simplify the Equation

Divide both sides of the equation by 2 to isolate the sine function: \[ \text{sin}(2\theta) = \frac{1}{2} \].

Solve for the Angle

Find the general solutions for \(\text{sin}(x) = \frac{1}{2}\). The solutions within one period \(0 \leq x < 2\text{π} \) are: \[ x = \frac{\text{π}}{6}, \frac{5\text{π}}{6} \]. Since \(x = 2\theta\), we have: \[ 2\theta = \frac{\text{π}}{6} \text{ or } 2\theta = \frac{5\text{π}}{6} \].

Solve for \(\theta\)

Divide each equation by 2 to solve for \(\theta\): \(\theta = \frac{\text{π}}{12} \) or \(\theta = \frac{5\text{π}}{12} \).

Consider the Full Period

Since \(\theta\) must be in the range \(0 \leq \theta < 2\text{π}\), add \(\text{π}\) to the solutions to get additional solutions in the full period: \( 2\theta = \frac{\text{π}}{6} + \text{π} \text{ or } 2\theta = \frac{5\text{π}}{6} + \text{π} \).Simplify the new angles: \[ 2\theta = \frac{7\text{π}}{6} \text{ and } 2\theta = \frac{11\text{π}}{6} \]. Now, divide by 2 again: \(\theta = \frac{7\text{π}}{12} \) or \(\theta = \frac{11\text{π}}{12} \).

List the Solutions

Combine all the solutions found to get the final \(\theta\): \[ \theta = \frac{\text{π}}{12} , \frac{5\text{π}}{12}, \frac{7\text{π}}{12}, \frac{11\text{π}}{12} \]. These are the angles that satisfy the given equation.

Key concepts

Double Angle Identity

In trigonometry, the double angle identity is a powerful tool used to simplify equations. The double angle identity for sine is written as: \[ \text{sin}(2\theta) = 2 \text{sin}(\theta) \text{cos}(\theta) \]. It allows us to express a sine function involving a double angle in terms of sine and cosine of a single angle.
This identity is particularly helpful when you need to convert a product of sine and cosine into a single sine term.
For the given problem, we start with the equation: \[ 4 \text{sin}(\theta) \text{cos}(\theta) = 1 \].
Using the double angle identity, we can rewrite it as: \[ 2 \text{sin}(2\theta) = 1 \].
Simplifying this, we get: \[ \text{sin}(2\theta) = \frac{1}{2} \].
This transformation simplifies solving the equation, and it turns an otherwise complex trigonometric equation into a more manageable form.

Sine Function

The sine function is one of the fundamental functions in trigonometry. It is crucial to know how it behaves and its key properties.
The sine function, denoted as \( \text{sin}(x) \), is periodic with a period of \( 2\pi \). This means that \( \text{sin}(x + 2\pi) = \text{sin}(x) \) for any angle \( x \).
The function features a range between -1 and 1. Mathematically: \[ -1 \leq \text{sin}(x) \leq 1 \]. It is maximum at \( \frac{\pi}{2} \) and minimum at \( \frac{3\pi}{2} \).
In the given problem, we use the key values where the sine function yields \( \frac{1}{2} \):

• \( \text{sin}( \frac{\pi}{6} ) = \frac{1}{2} \)
• \( \text{sin}( \frac{5\pi}{6} ) = \frac{1}{2} \)

Solving these within one period helps us find possible solutions for \( 2\theta \). Subsequently, we divide by 2 to back-solve for \( \theta \).

General Solutions for Sine

To find solutions for equations involving the sine function, you need to consider all possible angles, within the given range, that satisfy the equation.
For \( \text{sin}(x) = \frac{1}{2} \), the angle solutions in one period \( 0 \leq x < 2\pi \) are: \[ x = \frac{\pi}{6}, \frac{5\pi}{6} \].
When dealing with multiples or transformations of angles, you might need more solutions. For instance, here: \[ 2\theta = \frac{\pi}{6}, \frac{5\pi}{6} \], is obtained initially.
Considering the full period, you introduce solutions by adding \( \pi \): \[ 2\theta = \frac{\pi}{6} + \pi, \frac{5\pi}{6} + \pi \], which simplifies to: \[ 2\theta = \frac{7\pi}{6}, \frac{11\pi}{6} \].
You divide these by 2 to get: \( \theta = \frac{7\pi}{12}, \frac{11\pi}{12} \).
Combining all, the general solutions are:
\( \theta = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{7\pi}{12}, \frac{11\pi}{12} \).
This comprehensive approach ensures no solution within the specified range is missed.