Solve trigonometric equations

Given a particular value of a trigonometric function, we may wish to find the values of the domain variable that are mapped to that function value. Or, given two distinct functions defined on the same domain, we may ask what values of the domain variable make the two functions equal.

These situations are what is meant by the idea of solving trigonometric (and other) equations.  We are finding the values of the variable that make the equations true. 

Example 1

Given the function defined by $2\sin x$2sinx for $x\in[0,2\pi]$x∈[0,2π], find the values of $x$x for which the function attains the value $1$1.

The problem is to solve the equation $2\sin x=1$2sinx=1, giving the solutions that are between $0$0 and $2\pi$2π. The equation is rearranged to read $\sin x=\frac{1}{2}$sinx=12​ and we proceed by asking what values of $x$x produce this function value.

From previous work we recall that $\sin\frac{\pi}{6}=\frac{1}{2}$sinπ6​=12​ and consequently that $\sin\left(\pi-\frac{\pi}{6}\right)=\sin\frac{5\pi}{6}=\frac{1}{2}$sin(π−π6​)=sin5π6​=12​. The sine function is negative between $\pi$π and $2\pi$2π. So, these two are the only values that satisfy the equation.

We can give the solution as $x=\frac{\pi}{6}$x=π6​, $\frac{5\pi}{6}$5π6​.

If we had not known that $\sin\frac{\pi}{6}=\frac{1}{2}$sinπ6​=12​, it would have been necessary to use the inverse sine function. The steps are as follows:

$\sin x$sinx	$=$=	$\frac{1}{2}$12​
$x$x	$=$=	$\arcsin\frac{1}{2}$arcsin12​
	$\approx$≈	$0.5236$0.5236

The number $0.5236$0.5236 is obtained from a hand-held calculator or a desktop computer or from a printed table of sines. It is an approximation to the true value $\frac{\pi}{6}$π6​ which cannot be written down with a finite decimal string.

A calculator gives only one solution since the inverse sine function is defined only on the interval $[-\frac{\pi}{2},\frac{\pi}{2}]$[−π2​,π2​]. We subtract the calculator value from $\pi$π to obtain the second solution, $\pi-0.5236\approx2.6180$π−0.5236≈2.6180. 

Another way to consider the solution is to think of $2$2 functions $y=2\sin x$y=2sinx and $y=1$y=1.  To solve $2\sin x=1$2sinx=1, is the same as graphically finding the intersection of these two separate lines.  

You can see both solutions mentioned above here.  

 

Example 2

Solve $4\cos x+1=3$4cosx+1=3 for $x\in[0,2\pi]$x∈[0,2π].

The equation is rearranged to $\cos x=\frac{1}{2}$cosx=12​. In the domain $[0,2\pi]$[0,2π], we can expect to find a solution somewhere in the interval $[0,\frac{\pi}{2}]$[0,π2​] and another in the interval $[\frac{3\pi}{2},2\pi]$[3π2​,2π] as these are the intervals in which the cosine function is positive.

Again, from previous work, we know that $\cos x=\frac{1}{2}$cosx=12​ when $x=\frac{\pi}{3}$x=π3​. The next solution must occur when $x=2\pi-\frac{\pi}{3}$x=2π−π3​. That is, when $x=\frac{5\pi}{3}$x=5π3​.

Thus, the solutions are $x=\frac{\pi}{3},\frac{5\pi}{3}$x=π3​,5π3​.

The graphs for this can also be used to find the solution.  Graph both y=4\cos{x} + 1 and y=3 on the same set of axes.  The intersections of these curves are the solutions.  

 

Trigonometric equations do not always have easy solutions. It can be necessary to use a numerical method to obtain an approximate solution. 

Example 3

Two functions, defined by the expressions $\sin^2x$sin2x and $\cos\frac{x}{2}$cosx2​, are thought to be equal at certain points in the interval $[0,2\pi]$[0,2π]. Find the points if they exist.

We need to solve $\sin^2x=\cos\frac{x}{2}$sin2x=cosx2​. As a first step, we should make the arguments of the left-hand and right-hand functions the same by writing $\sin^2x$sin2x as a function of $\frac{x}{2}$x2​. 

To do this, we can use the double angle identity, $\sin2A\equiv2\sin A\cos A$sin2A≡2sinAcosA. Thus, we can put $\sin x\equiv2\sin\frac{x}{2}\cos\frac{x}{2}$sinx≡2sinx2​cosx2​ and so, the equation to be solved can be written

$\left(2\sin\frac{x}{2}\cos\frac{x}{2}\right)^2=\cos\frac{x}{2}$(2sinx2​cosx2​)2=cosx2​.

It is clear that both sides of this equation will equal zero when $\cos\frac{x}{2}=0$cosx2​=0. This occurs when $\frac{x}{2}=\frac{\pi}{2}\pm n\pi$x2​=π2​±nπ.
In the interval $[0,2\pi]$[0,2π], this is when $x=\pi$x=π.

We have one solution but there could be others. If the right- and left-hand functions are graphed together, we see that the graphs intersect twice between $0$0 and $\pi$π. However, it is not easy to find the solutions represented by these intersections by algebraic means.

Suppose $x\ne\pi$x≠π. Then $\cos\frac{x}{2}$cosx2​ is not zero and we can cancel it from both sides of the equation, leaving

$4\sin^2\frac{x}{2}\cos\frac{x}{2}=1$4sin2x2​cosx2​=1

Using the identity $\cos^2x+\sin^2x\equiv1$cos2x+sin2x≡1, we can transform the equation to a cubic polynomial equation in $\cos\frac{x}{2}$cosx2​.

$4\cos^3\frac{x}{2}-4\cos\frac{x}{2}+1=0$4cos3x2​−4cosx2​+1=0

Although not impossible, this cubic is difficult to solve,  

Another approach might be to write the equation as

$2\sin\frac{x}{2}\ \times\ 2\sin\frac{x}{2}\cos\frac{x}{2}=1$2sinx2​ × 2sinx2​cosx2​=1

That is, $2\sin\frac{x}{2}\sin x=1$2sinx2​sinx=1. This is still awkward, although it can lead to an iterative method that will provide one of the other solutions. Try this:

We can see from the graph shown above that the additional solutions should be approximately $x=1.15$x=1.15 and $x=2.6$x=2.6. Better solutions can be obtained using suitable software.

 

Worked Examples

Question 1

Question 2

Question 3