Trigonometric Equations

To solve a trigonometric equation, we need the following preliminary knowledge:

1. If \(\sin \theta = \sin \alpha\), then \(\theta=n\pi+(-1)^{n}\alpha\). Thus, if \(n\) is odd, \(\theta=(2m+1)\pi-\alpha\) and if \(n\) is even, \(\theta=2m\pi+\alpha\).

2. If \(\cos \theta = \cos \alpha\), then \(\theta = 2n\pi\pm\alpha\).

3. If \(\tan \theta = \tan\alpha\), then \(\theta=n\pi+\alpha\).

These hold true for integers \(n,m\).

Now on to solving equations. The general method of solving an equation is to convert it into the form of one ratio only. Then, using these results, we can obtain solutions.

Solving basic equations can be taken care of with the Trigonometric R method.

Consider the following example:

Solve the following equation: \[\]

\[\cos x - \sqrt{3}\sin x = 2.\]

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Using the R method, this equation can be converted into

\[\cos\left(x+\frac{\pi}{3}\right)=1.\]

Now, we know that \(\cos0^ \circ=1\). Hence we rewrite the equation as

\[\cos\left(x+\frac{\pi}{3}\right)=\cos0^\circ.\]

Using the above results, we have

\[\begin{align} x+\frac{\pi}{3} &= 2n\pi\pm0\\ x&=2n\pi-\frac{\pi}{3}. \ _\square \end{align}\]

Note: Sometimes the question may also provide a range for \(x\). In this case, we take various integral values for \(x\) and find the solutions.

But as the equations get harder, a variety of techniques come handy: the double and triple angle formulas, the sum to product formulas, etc. Sometimes the equation is converted into a form where the R method can be used, or sometimes we have quadratic equations in the field.

Let's take another example.

Find the general solution of the equation \(\cos2x-2\tan x+2=0.\)

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Note that \(\cos 2x = \frac{1-\tan^{2} x}{1+\tan^{2} x}.\)

Substituting this, the equation has only one variable: \(\tan x\). We evaluate and then factorise the expression to obtain

\[(\tan x - 1)\left(2\tan^{2} x + \tan x + 3\right)=0.\]

As the equation \(2\tan^{2} x + \tan x + 3 = 0\) has imaginary roots, we do not consider it. Hence, we have

\[\begin{align} \tan x - 1 &= 0\\ \tan x &= 1\\ x&=\frac{\pi}{4}+n\pi. \ _\square \end{align}\]

What are the solutions of the equation \[ \sin 2x = \frac{\sqrt{3}}{2} \] in the \(x\)-interval \( [0, 2\pi] ?\)

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Since \( 0 \leq 2x \leq 4\pi ,\) we have \[\begin{align} 2x&= \frac{\pi}{3}, \frac{2}{3} \pi, \frac{7}{3} \pi, \frac{8}{3} \pi\\ x&= \frac{\pi}{6}, \frac{\pi}{3}, \frac{7}{6} \pi, \frac{4}{3} \pi. \ _\square \end{align} \]

What are the solutions of \[ \sin \left(x-\frac{\pi}{3} \right) = -\frac{1}{2} \] in the \(x\)-interval \( [0, 2\pi] ?\)

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Since \( -\frac{\pi}{3} \leq x-\frac{\pi}{3} \leq \frac{5}{3} \pi ,\) we have \[\begin{align} x-\frac{\pi}{3} &=-\frac{\pi}{6}, \frac{7}{6} \pi\\ x &= \frac{\pi}{6}, \frac{3}{2} \pi. \ _\square \end{align}\]

What is the general solution of

\[ \tan x = 1 ?\]

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In the first period \( 0 \leq x < \pi,\) the solution for this equation is \(x= \frac{\pi}{4}.\) Since the period of \(\tan x\) is \(\pi,\) the general solution of the given equation is \[ x = n \pi + \frac{\pi}{4} . \ _\square\]

What is the general solution of

\[ \cos 2x +3\cos x -1 ?\]

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Since \( \cos 2x = 2\cos^2 x -1,\) we have \[ \begin{align} 2 \cos^2 x -1 + 3\cos x -1 &= 0 \\ (2\cos x -1)(\cos x +2) &= 0\\ \cos x &= \frac{1}{2}. \qquad (\text{since }\lvert \cos x \rvert \leq 1) \end{align} \] Since the period of \( \cos x \) is \( 2\pi,\) the general solution for the given equation is \[ x = 2n \pi \pm \frac{\pi}{3}. \ _\square \]

What is the general solution of

\[ \tan 2x - 3\tan x = 0 ?\]

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Since \( \tan 2x = \frac{2\tan x}{1-\tan^2 x} ,\) we have \[ \begin{align} \tan 2x - 3\tan x &= 0 \\ \frac{2\tan x}{1-\tan^2 x} -3\tan x &= 0 \\ 2\tan x -3\tan x \left(1-\tan^2 x\right) &= 0 \\ \tan x\left(3\tan^2 x -1\right) &= 0 \\ \Rightarrow \tan x &= 0, \pm \frac{1}{\sqrt{3}}. \end{align} \]

Since the period of \( \tan x\) is \(\pi,\) the general solution for the given equation is \[ x= n \pi, \text{ or } x=n \pi \pm \frac{\pi}{6}. \ _\square\]

What are the solutions of the following equation for \(0 \leq x \leq 2\pi:\) \[\]

\[ 2\cos^2 x - \sin x -1=0 ? \]

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Since \( \sin^2 + \cos^2 = 1 ,\) we rewrite the given equation to obtain

\[ \begin{align} 2 \cos^2 x-\sin x -1 &= 0 \\ \Rightarrow 2(1-\sin^2 x)-\sin x -1 &= 0 \\ \left(2\sin x - 1\right)(\sin x + 1) &=0 \\ \sin x &= \frac{1}{2}, -1. \end{align} \]

Since \(0 \leq x \leq 2\pi,\) for \( \sin x = \frac{1}{2}\) we have

\[ x = \frac{\pi}{6}, \frac{5}{6} \pi. \qquad (1) \]

For \( \sin x = -1,\) we have

\[ x = \frac{3}{2} \pi. \qquad (2) \]

Thus, from \( (1)\) and \( (2)\) we have

\[ x = \frac{\pi}{6}, \frac{5}{6} \pi, \frac{3}{2} \pi. \ _\square\]

What are the solutions of the following equation for \(0 \leq x \leq 2\pi:\) \[\]

\[ \sin^ 4 x + \sin^2 x -1 = \cos^4 x + \cos^2 x ?\]

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Using \(\sin^2 x + \cos^2 x = 1,\) we have

\[ \begin{align} \sin^ 4 x + \sin^2 x -1 &= \cos^4 x + \cos^2 x \\ \sin^4 x - \cos^4 x + \sin^2 x - \cos^2 x -1 &= 0 \\ \left(\sin^2 x + \cos^2 x\right)\left(\sin^2 x - \cos^2 x\right) + \sin^2x - \cos^2 x -1 &= 0 \\ 2\sin^2 x - 2\cos^2 x - 1 &= 0 \\ 2\sin^2 x - 2\left(1-\sin^2 x\right) -1 &= 0 \\ \Rightarrow \sin^2 x &= \frac{3}{4} \\ \sin x &= \pm \frac{\sqrt{3}}{2}. \end{align} \]

Since \(0 \leq x \leq 2\pi,\) for \( \sin x = \frac{\sqrt{3}}{2}\) we have

\[ x = \frac{\pi}{3}, \frac{2}{3} \pi . \qquad (1) \]

For \( \sin x = -\frac{\sqrt{3}}{2} ,\) we have

\[ x= \frac{4}{3} \pi, \frac{5}{3} \pi. \qquad (2) \]

Thus, from \( (1) \) and \( (2) \) the solutions are

\[x = \frac{\pi}{3}, \frac{2}{3} \pi, \frac{4}{3} \pi, \frac{5}{3} \pi. \ _\square\]

What are the solutions of the following equation for \(0 \leq x \leq 2\pi:\) \[\]

\[ \cos^2 x - \sin^2 2x = 0?\]

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We have

\[ \begin{align} \cos^2 x - \sin^2 2x &= 0 \\ \cos^2 x - 4\sin^2 x \cos^2 x &= 0 \qquad ( \text{ since } \sin 2x = 2\sin x \cos x ) \\ \cos^2x(1-4\sin^2 x) &= 0 \\ \Rightarrow \cos x &= 0, ~\sin x = \frac{1}{2}, - \frac{1}{2}. \end{align} \]

Since \(0 \leq x \leq 2\pi,\) for \( \cos x =0\) we have

\[ x = \frac{\pi}{2}, \frac{3}{2} \pi. \qquad (1) \]

For \( \sin x = \frac{1}{2},\) we have

\[ x = \frac{\pi}{6}, \frac{5}{6} \pi. \qquad (2) \]

For \( \sin x = -\frac{1}{2},\) we have

\[ x= \frac{7}{6} \pi, \frac{11}{6} \pi. \qquad (3) \]

Thus, from \( (1), (2) \) and \((3)\) the solutions are

\[ x= \frac{\pi}{2}, \frac{3}{2} \pi, \frac{\pi}{6}, \frac{5}{6} \pi, \frac{7}{6} \pi, \frac{11}{6} \pi. \ _\square\]

What are the values of \(\alpha\) in the interval \( [0, 2\pi]\) such that the quadratic equation \( x^2 + 2x + 2\cos \alpha=0\) has repeated roots?

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For the quadratic equation \( x^2 + 2x +2\cos \alpha = 0\) to have repeated roots, its discriminant \(D\) must be 0. Thus,

\[ \begin{align} \frac{D}{4} = 1-2\cos \alpha &= 0 \\ \cos \alpha &= \frac{1}{2} \\ \Rightarrow \alpha &= \frac{\pi}{3}, \frac{5}{3} \pi. \ _\square \end{align} \]

If a root of the equation \[\]

\[1-6\sin x = \sqrt{5-12\sin x} \]

is \( \alpha,\) what is the value of \( \sin \alpha \cos 2\alpha?\)

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Squaring both sides, we have

\[ \begin{align} (1-6\sin x)^2 &= \left(\sqrt{5-12\sin x}\right)^2 \\ 1-12\sin x + 36\sin^2 x &= 5 - 12 \sin x \\ \sin^2 x &= \frac{1}{9} \\ \Rightarrow \sin x &= \pm \frac{1}{3}. \end{align} \]

Since only \(\sin x = - \frac{1}{3} \) satisfies \( 1-6\sin x = \sqrt{5-12\sin x} ,\) we have \( \sin \alpha= -\frac{1}{3}.\)

Thus, \( \sin \alpha \cos 2\alpha \) is

\[ \begin{align} \sin \alpha \cos 2\alpha &= \sin \alpha\left(1-2\sin^2 \alpha\right) \\ &= -\frac{1}{3}\left( 1- \frac{2}{9} \right ) \\ &= -\frac{7}{27}. \ _\square \end{align} \]

If the roots of the quadratic equation \( 8x^2+7x+a=0\) are \( \sin \theta\) and \( \cos 2\theta,\) what is \(a?\)

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From Vieta's formulas, we have

\[ \begin{align} \sin \theta + \cos 2\theta &= -\frac{7}{8} &\qquad (1) \\ \sin \theta \cos 2\theta &= \frac{a}{8}. &\qquad (2) \end{align} \]

From \((1),\) we have

\[ \begin{align} \sin \theta + \cos 2\theta &= -\frac{7}{8} \\ 8(\sin \theta +1 - 2 \sin^2 \theta) + 7 &= 0 \\ (4\sin \theta +3)(4\sin \theta - 5) &= 0 \\ \Rightarrow \sin \theta &= -\frac{3}{4}, \frac{5}{4}. \end{align} \]

Since \(\lvert \sin \theta \rvert \leq 1,\) it follows that \( \sin \theta = -\frac{3}{4}.\)

Thus, we have

\[ \begin{align} \cos 2\theta &= 1-2\sin^2 \theta \\ &= 1-2\sin^2\left(-\frac{3}{4} \right) \\ &= -\frac{1}{8}. \end{align} \]

Substituting this into \((2)\) gives

\[ \begin{align} \sin \theta \cos 2 \theta &= \frac{a}{8} \\ \left( -\frac{3}{4} \right) \left( -\frac{1}{8} \right ) &= \frac{a}{8} \\ \Rightarrow a&= \frac{3}{4}. \ _\square \end{align} \]