Question

If \(2 \sin ^{2} \theta+2 \sin \theta \cos \theta=1\), then \(\theta\) can be (a) \(\frac{\pi}{8}-\frac{\mathrm{n} \pi}{2}, \mathrm{n} \in \mathrm{Z}\) (b) \(\frac{\pi}{2}-2 n \pi, n \in Z\) (c) \((2 n+1) \frac{\pi}{2}, n \in Z\) (d) \(\mathrm{n} \pi, \mathrm{n} \in \mathrm{Z}\)

Short answer

Answer: (a) \(\theta = \frac{\pi}{8} - \frac{n\pi}{2}\), \(n\in Z\).

Step-by-step solution

Simplify the given equation

The given equation is \(2 \sin ^{2} \theta+2 \sin \theta \cos \theta=1\). First, let's rewrite it using the double angle identity \(\sin 2\alpha = 2 \sin\alpha \cos\alpha\). The equation becomes \(\displaystyle 2 \sin^{2} \theta + \sin 2\theta = 1\).

Substitute a new variable for \(\sin\theta\)

Let's define a new variable \(x = \sin \theta\). This can make the equation more easily solvable. So now, our equation is \(2x^{2}+ \sin {2\theta} = 1\).

Convert back to sine

Getting back sine, we have \(2x^2+\sin(2\arcsin(x))=1\).

Use the sine double angle formula

Now we have \(2x^2+2x\sqrt{1-x^2}=1\).

Solve for x

Subtract 1 from both sides and divide the entire equation by 2: \(x^2 + x\sqrt{1-x^2}=0\). Factor out x: \(x(x+\sqrt{1-x^2})=0\). We get two solutions: \(x = 0\) and \(x = -\sqrt{1-x^2}\).

Solve for x and \(-\sqrt{1-x^2}\)

Firstly, solving \(x = 0\) gives us the solution \(\sin\theta = 0 \Rightarrow \theta = n\pi\), \(n\in Z\). Secondly, solving for \(x=-\sqrt{1-x^2}\), square both sides: \(x^2 = 1 - x^2\), which gives \(2x^2 = 1 \Rightarrow x = \pm\frac{1}{\sqrt{2}}\). This means that \(\sin\theta = \pm\frac{1}{\sqrt{2}} \Rightarrow \theta = \pm\frac{\pi}{4} + m\pi\), \(m \in Z\).

Combine the results

We found two possible solutions for \(\theta\): \(\theta = n\pi\) and \(\theta = \pm\frac{\pi}{4} + m\pi\). The first one matches option (d) and the second one can be rewritten as \(\theta = \frac{\pi}{8}\mp \frac{n\pi}{2}\). The correct answer is (a) \(\theta = \frac{\pi}{8} - \frac{n\pi}{2}\), \(n\in Z\).

Key concepts

Double Angle Formulas

Understanding double angle formulas is key to solving many trigonometric equations. These formulas express trigonometric functions of double angles, like \(2\alpha\), in terms of functions of \(\alpha\).
One of the most common is the double angle formula for sine: \(\sin 2\alpha = 2 \sin \alpha \cos \alpha\). This identity is useful in simplifying expressions where terms involve products of sine and cosine. For instance, given an equation like \(2 \sin \theta \cos \theta\), we can utilize this formula to rewrite it as \(\sin 2\theta\).
This simplification makes equations easier to handle and can help expose periodic solutions on sums of trigonometric functions. Mastering the double angle formulas can significantly streamline your mathematical problem-solving.

Sine Function

The sine function is one of the fundamental trigonometric functions and it defines the ratio of the length of the opposite side to the length of the hypotenuse in a right triangle. It is defined from -1 to 1 and is periodic with a period of \(2\pi\).
When solving trigonometric equations, identifying solutions often involves finding specific sine values for which the equation holds true. An important identity for sine that is frequently used is \(\sin^2 \theta + \cos^2 \theta = 1\), which relates sine and cosine values.
Additionally, understanding how the sine function repeats itself at regular intervals, or its periodicity, helps in finding all possible solutions \(\theta\) that satisfy a given equation. This repetition is what enables us to write solutions in the form \(\theta = n\pi\) or \(\theta = \pm \frac{\pi}{4} + m\pi\).

Periodic Solutions

In trigonometry, periodic solutions are solutions that repeat at regular intervals. Understanding this concept is vital in figuring out all possible solutions to trig equations. Since trigonometric functions like sine and cosine are periodic, they repeat their values in regular intervals. This means there are infinite solutions to \ such equations, typically expressed with a formula.
For instance, the sine and cosine functions have a period of \(2\pi\), which means the functions repeat every \(2\pi\) radians. When you solve for \(\theta\) in terms of \(n\), you capture every angle that satisfies the trigonometric equation.
In the equation discussed, solutions were expressed as \(\theta = \frac{\pi}{8} - \frac{n\pi}{2}\) and \(\theta = n\pi\). These formats let you understand that the solutions recur at regular intervals, providing a comprehensive understanding of all solutions involved.

Mathematical Problem Solving

Mathematical problem solving is an art that involves breaking down complex problems into simpler parts. This involves steps like substituting terms, using identities, and transforming equations. Let's take the problem-solving process for trigonometric equations as an example.
When you encounter an equation like \(2 \sin^2 \theta + 2 \sin \theta \cos \theta = 1\), a systematic approach involves using identities like the double angle formulas to simplify it. Recognizing that \(2 \sin \theta \cos \theta = \sin 2\theta\) allows you to reduce the terms and transform the problem into a familiar form.
Substitution can also be a helpful strategy. By using \(x = \sin \theta\), you exchange a complex trigonometric term for a simpler algebraic one, making it easier to solve. Once \(x\) is solved, you return to the original trigonometric terms to find the solutions for \(\theta\). Applying these methods supports you in achieving a deeper understanding and finding precise solutions in trigonometry.