Linear Trigonometric Equations: Videos & Practice Problems

In trigonometry, solving equations often involves understanding the unit circle and recognizing that certain trigonometric functions can yield multiple solutions. For instance, when given the equation sin(θ) = 1/2, it is essential to identify all angles within a specified interval, such as from 0 to 2π, where this condition holds true.

Starting with the unit circle, we find that θ = π/6 is one solution, as the sine of this angle equals 1/2. Additionally, in the second quadrant, θ = 5π/6 also satisfies the equation. Since sine values are negative in the third and fourth quadrants, these two angles represent all solutions within the interval of 0 to 2π.

However, to find all possible solutions without interval restrictions, we can extend our findings. For every complete rotation around the unit circle, which corresponds to adding 2πn (where n is any integer), we can generate additional solutions. For example, continuing from π/6 leads to 13π/6, and from 5π/6 leads to 17π/6. Thus, the general solution for sin(θ) = 1/2 can be expressed as:

θ = π/6 + 2πn and θ = 5π/6 + 2πn

Next, consider the equation cos(x) = -1. Here, we look for angles on the unit circle where the cosine value is -1. The only angle that satisfies this condition is x = π. To find all solutions, we again add 2πn to this angle, resulting in:

x = π + 2πn

By understanding how to navigate the unit circle and apply the concept of periodicity in trigonometric functions, we can effectively find all solutions to various trigonometric equations.

To solve the equation where the tangent of θ equals the square root of 3, we first identify the angles on the unit circle that satisfy this condition. The tangent function, defined as the ratio of the sine to the cosine (i.e., $\backslash (\backslash tan\backslash )(\backslash (\backslash theta\backslash ))\; =\; \backslash (\backslash frac\{\backslash sin(\backslash theta)\}\{\backslash cos(\backslash theta)\}\backslash )$), is positive in the first and third quadrants.

In the first quadrant, the angle $\backslash (\backslash frac\{\backslash pi\}\{3\}\backslash )$ has a tangent value of $\backslash (\backslash sqrt\{3\}\backslash )$. In the third quadrant, the angle $\backslash (\backslash frac\{4\backslash pi\}\{3\}\backslash )$ also yields a tangent of $\backslash (\backslash sqrt\{3\}\backslash )$. Thus, the two primary solutions are $\backslash (\backslash frac\{\backslash pi\}\{3\}\backslash )$ and $\backslash (\backslash frac\{4\backslash pi\}\{3\}\backslash )$.

To express all possible solutions, we add $2\backslash (\backslash pi\backslash )\; n$ to each of these angles, where $n$ is any integer, resulting in the general solutions:

• $\backslash (\backslash theta\backslash )\; =\; \backslash (\backslash frac\{\backslash pi\}\{3\}\backslash )\; +\; 2\backslash (\backslash pi\backslash )\; n$
• $\backslash (\backslash theta\backslash )\; =\; \backslash (\backslash frac\{4\backslash pi\}\{3\}\backslash )\; +\; 2\backslash (\backslash pi\backslash )\; n$

However, since $\backslash (\backslash frac\{\backslash pi\}\{3\}\backslash )$ and $\backslash (\backslash frac\{4\backslash pi\}\{3\}\backslash )$ are exactly $\backslash (\backslash pi\backslash )$ radians apart, we can simplify our solution further. By starting with $\backslash (\backslash frac\{\backslash pi\}\{3\}\backslash )$ and adding $\backslash (\backslash pi\backslash )\; n$, we can encompass all solutions. Therefore, the most simplified form of the solution is:

$\backslash (\backslash theta\backslash )\; =\; \backslash (\backslash frac\{\backslash pi\}\{3\}\backslash )\; +\; \backslash (\backslash pi\backslash )\; n$

This simplification is specific to the tangent function, as it exhibits periodicity with a period of $\backslash (\backslash pi\backslash )$. Thus, the final answer for the equation $\backslash (\backslash tan\backslash )(\backslash (\backslash theta\backslash ))\; =\; \backslash (\backslash sqrt\{3\}\backslash )$ is:

$\backslash (\backslash theta\backslash )\; =\; \backslash (\backslash frac\{\backslash pi\}\{3\}\backslash )\; +\; \backslash (\backslash pi\backslash )\; n$

To solve the equation where the sine of theta equals \( \frac{1}{2} \), we start by identifying the angles on the unit circle where this condition holds true. The sine function is positive in the first and second quadrants. Specifically, we find that \( \theta = \frac{\pi}{6} \) and \( \theta = \frac{5\pi}{6} \) are the angles corresponding to this sine value.

Next, we convert these radian measures into degrees. The conversion from radians to degrees can be achieved using the formula:

\( \text{Degrees} = \text{Radians} \times \frac{180}{\pi} \)

Applying this to our angles:

For \( \frac{\pi}{6} \):

\( \frac{\pi}{6} \times \frac{180}{\pi} = \frac{180}{6} = 30 \text{ degrees} \)

For \( \frac{5\pi}{6} \):

\( \frac{5\pi}{6} \times \frac{180}{\pi} = \frac{5 \times 180}{6} = 150 \text{ degrees} \)

Now that we have the angles in degrees, we need to express all possible solutions. Since the sine function is periodic, we can add full rotations to our solutions. In degrees, a full rotation is \( 360 \) degrees. Therefore, we express the general solutions as:

\( \theta = 30 + 360n \) and \( \theta = 150 + 360n \)

where \( n \) is any integer. This notation accounts for all angles that satisfy the original equation.

To solve linear trigonometric equations, we can apply basic algebraic techniques similar to those used in solving linear equations. For example, consider the equation:

4 sin(θ) - 3 = 1.

To solve this, we first isolate the trigonometric function, treating sin(θ) as a variable. We can rearrange the equation by adding 3 to both sides:

4 sin(θ) = 4.

Next, we divide both sides by 4:

sin(θ) = 1.

Now, we can use the unit circle to find the angle θ. The sine function equals 1 at:

θ = π/2.

This process illustrates how we can simplify more complex trigonometric equations back to basic forms that we can solve using known values from the unit circle.

For a more complex example, consider the equation:

-2 cos(θ) + √3 = 0.

Again, we start by isolating the cosine function. First, we subtract √3 from both sides:

-2 cos(θ) = -√3.

Next, we divide by -2:

cos(θ) = √3/2.

Now, we look for angles where the cosine equals √3/2. From the unit circle, we find:

θ = π/6 and θ = 11π/6.

Since the problem specifies finding solutions within the interval [0, 2π], we note that these two angles are valid solutions. If the domain were unrestricted, we would add 2πn to each solution, where n is any integer, but in this case, we are limited to the specified interval.

Finally, we confirm that θ is already isolated in the cosine function, completing our solution process. Thus, the solutions to the equation are:

θ = π/6 and θ = 11π/6.

By following these steps—isolating the trigonometric function, identifying solutions on the unit circle, and considering the domain—we can effectively solve linear trigonometric equations.

To solve the equation \(4 \cos(2\theta + \pi) + 8 = 12\), we begin by isolating the trigonometric function. First, we subtract 8 from both sides, resulting in:

\(4 \cos(2\theta + \pi) = 4\)

Next, we divide both sides by 4 to isolate the cosine function:

\(\cos(2\theta + \pi) = 1\)

Now, we need to determine the angles for which the cosine equals 1. Referring to the unit circle, we find that the cosine function equals 1 at:

\(2\theta + \pi = 0 + 2\pi n\)

where \(n\) is any integer. This equation represents all solutions since cosine is periodic with a period of \(2\pi\). To isolate \(\theta\), we first subtract \(\pi\) from both sides:

\(2\theta = 2\pi n - \pi\)

Next, we divide by 2:

\(\theta = \pi n - \frac{\pi}{2}\)

This final expression, \(\theta = \pi n - \frac{\pi}{2}\), represents all solutions to the original equation. By following these systematic steps, we can confidently arrive at the correct answer.