Some trigonometric equations are quadratic in a particular trigonometric function, in that they involve a trigonometric function being squared. In this way, many methods for solving quadratic equations can be used to solve trig equations.
Subsection14.4.1Notation \(\sin^2{x}\) and Other Powers of a Trigonometric Function
Sometimes, we want to write the square of a trig function. For example, \(\sin{x} \cdot \sin{x}\text{,}\) or \(\brac{\sin{x}}^2\text{.}\) It is common to write this more simply as \(\sin^2{x}\text{.}\)
This convention saves space and avoids writing so many brackets with expressions that have many powers involving trigonometric functions. This notation applies in general to all trigonometric functions,
This notation often confuses students when first learning it. In particular, the exponent (2 in this case) applies to the entire trigonometric function, not the angle \(x\text{.}\) In other words,
\begin{align*}
\underbrace{\sin^2{x}}_{\text{Sine of } x \text{, then squared}} \amp \neq \underbrace{\sin{\brac{x^2}}}_{\text{Take the sine of } x^2}
\end{align*}
If you like, whenever you see a squared trig function, you can first write it with brackets, before starting with the problem.
Consider \(2\sin^2{x} - 1 = 0\text{.}\) To solve this, you can think of replacing \(\sin{x}\) with a single variable \(u\) (\(u = \sin{x}\)), to get \(2u^2-1=0\text{.}\) To solve this, you would isolate for \(u^2\text{,}\) and then take the square root of both sides.
More precisely, if \(p(x) = 0\) is a quadratic equation, then the equation can be put into the form \(p(\sin{x}) = 0\) (or another trigonometric function).
Answer: \(x = 0, \pi, \frac{\pi}{2}\text{.}\) General solution: \(x = 0 + 2\pi n, x = \pi + 2\pi n, x = \frac{\pi}{2} + 2\pi n, n \in \mathbb{I}\) OR \(x = n\pi\) or \(x = \frac{\pi}{2} + 2\pi n, n \in \mathbb{I}\text{.}\)
Answer: \(x = \frac{\pi}{3}, \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{3}\text{.}\) General solution: \(x = \frac{\pi}{3} + 2\pi n, x = \frac{5\pi}{3} + 2\pi n, x = \frac{\pi}{2} + 2\pi n, x = \frac{3\pi}{2} + 2\pi n, n \in \mathbb{I}\) OR \(x = \frac{\pi}{2} + \pi n\) or \(x = \frac{\pi}{3} + 2\pi n\) or \(x = \frac{5\pi}{3} + 2\pi n, n \in \mathbb{I}\text{.}\)
Answer: \(x = \frac{\pi}{6}, \frac{11\pi}{6}, \frac{\pi}{2}, \frac{3\pi}{2}\text{.}\) General solution: \(x = \frac{\pi}{6} + 2\pi n, x = \frac{11\pi}{6} + 2\pi n, x = \frac{\pi}{2} + 2\pi n, x = \frac{3\pi}{2} + 2\pi n, n \in \mathbb{I}\) OR \(x = \frac{\pi}{2} + \pi n\) or \(x = \frac{\pi}{6} + 2\pi n\) or \(x = \frac{11\pi}{6} + 2\pi n, n \in \mathbb{I}\text{.}\)
Answer: \(x = 0, \pi, \frac{\pi}{6}, \frac{5\pi}{6}\text{.}\) General solution: \(x = 0 + 2\pi n, x = \pi + 2\pi n, x = \frac{\pi}{6} + 2\pi n, x = \frac{5\pi}{6} + 2\pi n, n \in \mathbb{I}\) OR \(x = n\pi\) or \(x = \frac{\pi}{6} + 2\pi n\) or \(x = \frac{5\pi}{6} + 2\pi n, n \in \mathbb{I}\text{.}\)
Answer: \(x = 0, \pi, \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}\text{.}\) General solution: \(x = 0 + 2\pi n, x = \pi + 2\pi n, x = \frac{\pi}{3} + 2\pi n, x = \frac{2\pi}{3} + 2\pi n, x = \frac{4\pi}{3} + 2\pi n, x = \frac{5\pi}{3} + 2\pi n, n \in \mathbb{I}\) OR \(x = n\pi\) or \(x = \frac{\pi}{3} + \pi n\) or \(x = \frac{2\pi}{3} + \pi n, n \in \mathbb{I}\text{.}\)
Note that you can’t divide by \(\sin{x}\) (or any other trig function), because that assumes that \(\sin{x} \neq 0\text{,}\) when in fact \(\sin{x}\) can equal 0. This is just like how with the equation \(x^2 - 2x = 0\text{,}\) you can’t divide by \(x\) and instead have to factor it out.
Exercise Group14.4.3.Trigonometric equations in quadratic form.
Solve each equation for \(0 \leq x \lt 2\pi\text{,}\) and give the general solution. Give exact values for special angles, otherwise give a decimal answer rounded to the nearest hundredth.
Answer: \(x = \frac{\pi}{3}, \pi, \frac{5\pi}{3}\text{.}\) General solution: \(x = \frac{\pi}{3} + 2\pi n, x = \pi + 2\pi n, x = \frac{5\pi}{3} + 2\pi n, n \in \mathbb{I}\text{.}\)
Answer: \(x = \frac{\pi}{6}, \frac{5\pi}{6}\text{.}\) General solution: \(x = \frac{\pi}{6} + 2\pi n, x = \frac{5\pi}{6} + 2\pi n, n \in \mathbb{I}\text{.}\)
Answer: \(x \approx 0.92, 2.22, 3.82, 5.60\text{.}\) General solution: \(x \approx 0.92 + 2\pi n, x \approx 2.22 + 2\pi n, x \approx 3.82 + 2\pi n, x \approx 5.60 + 2\pi n, n \in \mathbb{I}\text{.}\)
Answer: \(x \approx 1.23, \pi, 5.05\text{.}\) General solution: \(x \approx 1.23 + 2\pi n, x = \pi + 2\pi n, x \approx 5.05 + 2\pi n, n \in \mathbb{I}\text{.}\)
Answer: \(x = 0, \frac{\pi}{3}, \frac{5\pi}{3}\text{.}\) General solution: \(x = 2\pi n, x = \frac{\pi}{3} + 2\pi n, x = \frac{5\pi}{3} + 2\pi n, n \in \mathbb{I}\text{.}\)
Answer: \(x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}\text{.}\) General solution: \(x = \frac{\pi}{6} + 2\pi n, x = \frac{5\pi}{6} + 2\pi n, x = \frac{3\pi}{2} + 2\pi n, n \in \mathbb{I}\text{.}\)
Answer: \(x = \frac{\pi}{3}, \pi, \frac{5\pi}{3}\text{.}\) General solution: \(x = \frac{\pi}{3} + 2\pi n, x = \pi + 2\pi n, x = \frac{5\pi}{3} + 2\pi n, n \in \mathbb{I}\text{.}\)
Answer: \(x = \frac{\pi}{4}, \frac{5\pi}{4}, 2.82, 5.96\text{.}\) General solution: \(x = \frac{\pi}{4} + \pi n, x = 2.82 + \pi n, n \in \mathbb{I}\text{.}\)
Answer: \(x = 0.73, 2.41, 3.99, 5.44\text{.}\) General solution: \(x = 0.73 + 2\pi n, x = 2.41 + 2\pi n, x = 3.99 + 2\pi n, x = 5.44 + 2\pi n, n \in \mathbb{I}\text{.}\)
Solve each equation for \(0 \leq x \lt 2\pi\text{,}\) and give the general solution. Give exact values for special angles, otherwise give a decimal answer rounded to the nearest hundredth.
Answer: \(x = \frac{\pi}{3}, \frac{5\pi}{3}\text{.}\) General solution: \(x = \frac{\pi}{3} + 2\pi n, x = \frac{5\pi}{3} + 2\pi n, n \in \mathbb{I}\text{.}\)
Answer: \(x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}\text{.}\) General solution: \(x = \frac{\pi}{6} + 2\pi n, x = \frac{5\pi}{6} + 2\pi n, x = \frac{7\pi}{6} + 2\pi n, x = \frac{11\pi}{6} + 2\pi n, n \in \mathbb{I}\text{.}\)
Answer: \(x = 0, \pi, 0.96, 5.33\text{.}\) General solution: \(x = 0 + 2\pi n, x = \pi + 2\pi n, x = 0.96 + 2\pi n, x = 5.33 + 2\pi n, n \in \mathbb{I}\text{.}\)
Answer: \(x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{\pi}{2}, \frac{5\pi}{4}, \frac{7\pi}{4}, \frac{3\pi}{2}\text{.}\) General solution: \(x = \frac{\pi}{4} + \pi n, x = \frac{3\pi}{4} + \pi n, x = \frac{\pi}{2} + \pi n, n \in \mathbb{I}\text{.}\)
Answer: \(x = \frac{\pi}{3}, \frac{5\pi}{3}\text{.}\) General solution: \(x = \frac{\pi}{3} + 2\pi n, x = \frac{5\pi}{3} + 2\pi n, n \in \mathbb{I}\text{.}\)
The equation \(a \cos^2{x} + b \cos{x} - 1 = 0\) has solutions \(\frac{\pi}{3}, \pi\text{,}\) and \(\frac{5\pi}{3}\) on the interval \(0 \leq x \lt 2\pi\text{.}\) Find the values of \(a\) and \(b\text{.}\)
The equation \(\cot^2{x} - b \cot{x} + c = 0\) has solutions \(\frac{\pi}{6}, \frac{\pi}{4}, \frac{7\pi}{6}, \frac{5\pi}{4}\) on the interval \(0 \leq x \lt 2\pi\text{.}\) Find the values of \(b\) and \(c\text{.}\)
Solve \(2\cot{x} + \sec^2{x} = 0\text{.}\)Hint: write in terms of \(\tan{x}\) only, to get \(u^3 + u - 2 = 0\text{,}\) and factor using synthetic division. Answer: \(x = \frac{3\pi}{4}, \frac{7\pi}{4}\text{.}\) General solution: \(x = \frac{3\pi}{4} + \pi n, n \in \mathbb{I}\text{.}\)
