Question

Solve each equation for all nonnegative values of \(x\) less than \(360^{\circ} .\) Do some by calculator. $$3 \sin (x / 2)-1=2 \sin ^{2}(x / 2)$$

Short answer

The nonnegative values of \(x\) less than \(360^\circ\) that satisfy the equation are \(x = 60^\circ\), \(x = 180^\circ\), and \(x = 300^\circ\).

Step-by-step solution

Isolate Sin(x/2)

To solve the trigonometric equation, first rearrange the terms to isolate the \(\sin(x/2)\) term. This can be done by adding 1 to both sides, resulting in \(3\sin(x/2) = 2\sin^2(x/2) + 1\).

Rewrite the Equation

Subtract \(2\sin^2(x/2)\) from both sides to get a quadratic form in terms of \(\sin(x/2)\), leading to \(0 = 2\sin^2(x/2) - 3\sin(x/2) + 1\).

Factor the Quadratic Equation

Factor the quadratic equation \(2\sin^2(x/2) - 3\sin(x/2) + 1 = 0\) to find two binomials that multiply to give the quadratic expression. The factored form is \((2\sin(x/2) - 1)(\sin(x/2) - 1) = 0\).

Find Solutions for Sin(x/2)

Set each factor equal to zero to find solutions for \(\sin(x/2)\). This gives us two equations: \(2\sin(x/2) - 1 = 0\) and \(\sin(x/2) - 1 = 0\).

Solve the First Equation

Solve \(2\sin(x/2) - 1 = 0\) for \(\sin(x/2)\). Adding 1 to both sides and then dividing by 2 yields \(\sin(x/2) = 1/2\).

Solve the Second Equation

Solve \(\sin(x/2) - 1 = 0\) for \(\sin(x/2)\). Adding 1 to both sides gives \(\sin(x/2) = 1\).

Find Angles corresponding to Sin(x/2)

Use a calculator or unit circle to find angles \(x/2\) where the sine function equals 1/2 and 1. The solutions for \(\sin(x/2) = 1/2\) are \(x/2 = 30^\circ\) and \(x/2 = 150^\circ\). The solution for \(\sin(x/2) = 1\) is \(x/2 = 90^\circ\).

Calculate Corresponding Angles for x

Multiply each solution for \(x/2\) by 2 to find the corresponding angles for \(x\). The solutions are \(x = 60^\circ\), \(x = 300^\circ\), and \(x = 180^\circ\).

Check for nonnegative angles less than 360 degrees

Verify that all solutions for \(x\) are nonnegative and less than \(360^\circ\). All obtained angles satisfy these conditions, so no further restrictions apply to the solutions.

Key concepts

Trigonometric Identities

Understanding trigonometric identities is essential when solving trigonometric equations like the one given in the exercise. These identities are equations that are true for all values of the variables involved. For the given problem, knowledge of the Pythagorean identity, \(\sin^2(x) + \cos^2(x) = 1\), is not directly applicable, but having a firm grasp of such relations can help in more complex scenarios. There are numerous other trigonometric identities, such as the angle sum and difference identities, double angle formulas, and half-angle formulas, which can transform trigonometric expressions to a form that is easier to solve.

When approaching a problem, recognizing patterns that match these identities can greatly simplify the process. In the absence of a direct identity application, students can still employ their conceptual understanding to manipulate and rearrange the equation, as demonstrated in Step 1 of our solution, where isolating the \(\sin(x/2)\) term prepares the equation to be treated as a quadratic, which is then factored as shown in Steps 2 and 3.

Factoring Quadratic Equations

Factoring is a powerful technique employed to solve quadratic equations, which are of the form \(ax^2 + bx + c = 0\). The solution steps show how the original trigonometric equation was rewritten in a quadratic form involving \(\sin(x/2)\). When factoring, the goal is to decompose the quadratic expression into a product of two binomials. The key is to find two numbers that multiply to give the product of \(a\) and \(c\) and add to \(b\).

In our example, the equation \(2\sin^2(x/2) - 3\sin(x/2) + 1 = 0\) is factored into \(2\sin(x/2) - 1)(\sin(x/2) - 1) = 0\), as depicted in Step 3. Each factor is then set to zero to find possible solutions for \(\sin(x/2)\). This process parallels factoring a standard quadratic equation, with \(\sin(x/2)\) effectively playing the role of \(x\) in our algebraic expression. After factoring, as done in Step 4 and onward, we solve each resulting equation to find the specific values that satisfy the original equation.

Unit Circle

The unit circle is an invaluable tool in trigonometry because it helps to visualize sine, cosine, and tangent functions over different angles. For trigonometric equations, the unit circle allows one to find angle measures when the value of a trigonometric function is known.

In the last step of our solution, the unit circle was implicitly used to find angles corresponding to given sine values. For instance, when solving for angles where \(\sin(x/2) = 1/2\), the unit circle indicates that the angles in the first and second quadrants with this sine value are \(30^\circ\) and \(150^\circ\) respectively. Similarly, \(\sin(x/2) = 1\) only occurs at \(90^\circ\) on the unit circle. Through the use of the unit circle or appropriate technology, these angles provide the necessary \(x/2\) values that can then be multiplied by 2 to obtain the solutions for \(x\), as demonstrated in Step 8.

The unit circle epitomizes the cyclical nature of trigonometric functions, illustrating why certain sine and cosine values recur at specific intervals. This cyclic property is vital when solving for angles within a given domain, such as nonnegative values less than \(360^\circ\), since it allows us to identify all possible solutions within that range.