How many solutions do these simultaneous trig equations have?

Question

The simultaneous equations in $x$ , $y$ , $(\cos\theta)x - (\sin\theta)y = 2$ $(\sin\theta)x + (\cos\theta)y = 1$ are solvable

for all values of $\theta$ in the range $0 \leq \theta < 2\pi$ ;

except for one value of $\theta$ in the range $0 \leq \theta < 2\pi$ ;

except for two values of $\theta$ in the range $0 \leq \theta < 2\pi$ ;

except for three values of $\theta$ in the range $0 \leq \theta < 2\pi$ .

Suggestion

In this applet, the blue line is $x\sin \theta + y\cos \theta = 1$ , and the red line is $x\cos \theta - y\sin \theta= 2$ .

How will each of the lines change as we change $\theta$ ?

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/d7zWPYq7/width/600/height/400/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

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The intersection point, $A$ , represents the solution to the simultaneous equations. For which values of $\theta$ is the point $A$ uniquely defined?

Solution

Method 1

We write $c$ for $\cos\theta$ and $s$ for $\sin\theta$ , so the equations become $cx-sy=2, \qquad sx+cy=1;$ We multiply by $c$ and $s$ respectively and then add to eliminate $y$ , giving $c^2x-csy = 2c, \qquad s^2x+csy = s,$ so $2c+s = (c^2+s^2)x = x.$ Similarly, we can eliminate $x$ , since $csx - s^2y = 2s, \qquad csx + c^2y = c$ so $c - 2s = (s^2+c^2)y = y.$ Hence we have $x = 2\cos\theta + \sin\theta$ and $y = \cos\theta - 2\sin\theta$ , which are valid solutions for any value of $\theta$ , and the answer is (a).

Method 2

For a given $\theta$ , each equation represents a straight line.

So we will have exactly one solution unless the lines coincide, or are parallel, that is, unless they have the same gradient.

The gradient of the first line is $\dfrac{\cos \theta}{\sin \theta}$ , while the gradient of the second is $-\dfrac{\sin \theta}{\cos \theta}$ .

These can only be equal if $\dfrac{\cos \theta}{\sin \theta}=-\dfrac{\sin \theta}{\cos \theta}$ , which is true if and only if $\cos^2 \theta + \sin^2 \theta = 0$ , which is never true.

So there is always exactly one solution for any given $\theta$ , and the answer is (a).

Method 3

This requires compound angle formulae

If we square and add the two equations, we get $x^2 + y^2 = 5$ . Thus any solution must be of the form $x = \sqrt{5}\cos\alpha, y = \sqrt{5}\sin\alpha, \quad 0\leq \alpha < 2\pi.$

Substituting this back into our equations (since squaring may have introduced extra solutions), we have $\cos(\theta + \alpha) = \dfrac{2}{\sqrt{5}}, \sin(\theta + \alpha) = \dfrac{1}{\sqrt{5}}.$

Thus $\theta + \alpha$ is in the first quadrant, and for any $\theta$ there is exactly one solution for $\alpha$ , and the answer is (a).