Trigonometric Equations – Examples with Answers

We can write as follows:

$$\tan^2x-3=0\Rightarrow\tan^2x=3$$

$$\tan x=\pm\sqrt{3}$$

The unknown is isolated by using the corresponding inverse trigonometric function:

$$x =\arctan (\pm\sqrt{3})$$

That is, we must find all the angles whose tangent is $latex \pm\sqrt{3}$ and which at the same time belong to the interval $latex 0\leq x\leq 360º$. The equation has four of these solutions:

• $latex x_1 = 60º$
• $latex x_2 = 120º$
• $latex x_3 = 240º$
• $latex x_4 = 300º$

The reader can easily verify that:

$latex \tan 60º=\sqrt {3}=\tan 240º$

Whereas:

$latex \tan 120º=-\sqrt {3}=\tan 300º$

$$2\cos x -6=-4\Rightarrow 2\cos x=2$$

$$\cos x=1$$

Therefore:

$$x=\arccos 1$$

The solution of this equation in the requested interval is:

$latex x = 0º$

In this equation, we see that the variable $latex x$ is in the argument of cosine and sine, so the first step is to write the equation in terms of a single trigonometric function, for which a trigonometric identity relating them is required.

This identity is: $latex \cos^2 x + \sin^2 x = 1$.

We can now rewrite it to express cosine in terms of sine:

$latex \cos^2 x = 1- \sin^2 x$

And this is substituted into the original equation:

$$2\cos^2x-\sin x-1=0\Rightarrow 2(1- \sin^2 x)-\sin x-1=0$$

Leading to:

$$2- 2\sin^2 x-\sin x-1=0$$

$$- 2\sin^2 x-\sin x+1=0$$

Multiplying both sides by $latex -1$, we have:

$$2\sin^2 x+\sin x-1=0$$

Now the following variable substitution is made:

$latex y=\sin x $

And the equation is transformed into:

$$2y^2 +y-1=0$$

This is a second-degree equation of $latex y$ that is easily solved by factoring:

$$2y^2 +y-1=(2y-1)(y+1)=0$$

The roots of this equation are:

$latex y_1 =\dfrac{1}{2}$

$latex y_2 =-1$

The next step is to substitute back, which leads to two simple equations:

Equation 1

$$\sin x =\dfrac{1}{2}\Rightarrow x=\arcsin \left(\dfrac{1}{2}\right)$$

In the interval $latex 0\leq x\leq 360º$ there are two angles whose sine equal $latex \dfrac{1}{2}$, and they are:

• $latex x_1 =30º$
• $latex x_2 = 150º$

Equation 2

$$\sin x =-1\Rightarrow x=\arcsin (-1)$$

In the interval $latex 0\leq x\leq 360º$ there is a single angle whose sine equals $latex -1$, and it is:

• $latex x_3=270º$

As in the previous example, we need to have only one trigonometric function in the equation, and since tangent and secant appear here, we use the identity:

$latex \sec^2 x = 1+\tan ^2x$

Substituting in the original equation, it becomes:

$$\tan x+\sec^2 x – 3 = 0\Rightarrow \tan x+(1+\tan ^2x) – 3 = 0 $$

Therefore, the equation is:

$$ \tan ^2x+\tan x- 2 = 0 $$

This equation is also transformed into a quadratic equation by substituting $latex y=\tan x $ in:

$$y^2 +y-2=0$$

Easily solved by factoring:

$$y^2 +y-2=(y+2)(y-1)=0$$

And the roots are:

$latex y_1 =-2$

$latex y_2 =1$

Substituting back, this leads to two simple equations:

Equation 1

$latex \tan x =-2$

And the solution is:

$latex x = \arctan (-2)$

• $latex x_1=-63,4º; 116.6º; -243.40º;…$

Note that the statement did not specify an interval for the solution. When the solutions of a trigonometric equation lie in the interval $latex 0\leq x\leq 360º$, they are called main solutions.

But, due to the periodicity of trigonometric functions, there are an infinite number of solutions to $latex x = \arctan (-2)$ , which can be given in a general and compact form:

• $latex x_1 = k\cdot180º-63.4º$

Where $latex k$ is an integer, that is, $latex k = \pm 1;\pm 2; \pm 3; \pm 4…$

Equation 2

$latex \tan x =1$

Where it follows that:

$latex x = \arctan (1)$

• $latex x_2= 45º; 225º; -135º;…$
• $latex x_2 = k\cdot 180º+45º$

If preferred, the solution can be given in radians, as follows:

• $latex x_2 =k\pi+\dfrac{\pi}{4}$

Where $latex \dfrac{\pi}{4}=45º$, or:

• $latex x_2=\dfrac{(4k+1)\pi}{4}$

Where $latex k =0; \pm 1;\pm 2; \pm 3; \pm 4…$

Since the argument of sine and cosine are not the same, we use a trigonometric identity that relates $latex \sin 2x$ to $latex \sin x$ or $latex \cos x$, so that the argument is $ latex x$.

This identity is that of the double angle:

$latex \sin 2x =2\sin x\cos x$

Which, when substituted into the original equation, gives as this result:

$$\sin 2x +\cos x = 0\Rightarrow 2\sin x\cos x+\cos x=0$$

The equation obtained is factored to find the solutions, leaving:

$$\cos x(2\sin x+1)=0$$

From there, two simpler equations are obtained:

Equation 1

$latex \cos x=0$

$latex x =\arccos 0$

Whose solutions, in the interval indicated in the statement, are:

• $latex x_1 =90º=\dfrac{\pi}{2}$
• $latex x_2 =270º=\dfrac{3\pi}{2}$

Equation 2

$latex 2\sin x+1=0\Rightarrow 2\sin x=-1$

$latex \sin x=-\dfrac{1}{2}$

$latex x=\arcsin\left(-\dfrac{1}{2}\right)$

Which gives us two solutions in the indicated interval:

• $latex x_3 =210º=\dfrac{7\pi}{6}$
• $latex x_4 =330º=\dfrac{11\pi}{6}$

This trigonometric equation is of the form:

$latex a\sin x + b\cos x = 0$

Which is easily solved by dividing both sides of the equality by $latex \cos x$, to leave everything in terms of $latex \tan x$:

$$3\sin x-\sqrt{3}\cos x=\frac{3\sin x-\sqrt{3}\cos x}{\cos x}=0$$

This equation becomes:

$$3\tan x-\sqrt{3}=0$$

Therefore:

$$3\tan x=\sqrt{3}\Rightarrow\tan x=\frac{\sqrt{3}}{3}$$

$$x = \arctan \dfrac{\sqrt{3}}{3}$$

Since the statement did not specify an interval, it means that there are infinite solutions:

• $latex x_1 = 30º, 210º, 390º, 570º…$

And its general form is:

• $latex x = k\pi+\dfrac{\pi}{6}$

Or:

• $latex x = k\cdot 180º+30º$

Where $latex k = 0, \pm 1, \pm 2, \pm 3…$

To solve this equation, which contains an independent term other than $latex 0$, the following change of variable is used:

$$t=\tan \left(\dfrac{x}{2}\right)$$

Thus, sine and cosine are expressed as follows:

$$\cos x=\dfrac{1-t^2}{1+t^2}$$

$$\sin x=\dfrac{2t}{1+t^2}$$

This change of variable is known as the universal substitution or Weierstrass substitution. Then, the original equation looks like this:

$$4\sin x+3\cos x = 3\Rightarrow 4\left(\frac{2t}{1+t^2}\right) +3\left(\frac{1-t^2}{1+t^2}\right)=3$$

$$\frac{8t+3-3t^2}{1+t^2}=3\Rightarrow 8t+3-3t^2=3+3t^2$$

Regrouping the terms, we obtain a quadratic equation in $latex t$:

$$-6t^2+8t=0$$

$$3t^2-4t=0$$

Which is solved by factoring:

$$3t^2-4t=0\Rightarrow t(3t-4)=0$$

The solutions are:

$latex t_1 =0$

$latex t_2=\dfrac{4}{3}$

This leads to two simple equations, by substituting back:

Equation 1

$$\tan \left(\dfrac{x}{2}\right)=0$$

Therefore:

$latex\dfrac{x}{2}=\arctan 0$

$latex \dfrac{x}{2}=0, \pm\pi, \pm 2\pi, \pm 3\pi,…$

$latex x=0, \pm 2\pi, \pm 4\pi, \pm 6\pi,…$

$latex x=2k\pi=2k\cdot 360º$

Where $latex k$ is an integer.

Equation 2

$$\tan \left(\dfrac{x}{2}\right)=\dfrac{4}{3}$$

We deduce that:

$latex \dfrac{x}{2}=\arctan \left(\dfrac{4}{3}\right)=53.1º$

Therefore, the main solution, which is in the interval $latex 0\leq x\leq 360º$, is:

$latex x=106.2º$

Furthermore, the angles $latex -253.8º, 466.2º,…$ are also solutions of the proposed equation, since the tangent function is periodic, therefore, in general form, the solution is given by:

$latex x = k\cdot 360º+ 106.2º$

Where $latex k = 0,\pm 1, \pm 2, \pm 3…$, that is, $latex k$ is an integer.

If the Pythagorean identity is used:

$latex \sin^2x+\cos^2x$

$latex \cos^2x=1-\sin^2x$

Then:

$$3\sin^2x-\cos^2x =0\Rightarrow 3\sin^2x-(1-\sin^2x )=0 $$

Therefore:

$$4\sin^2x-1 =0$$

$$\sin^2x=\dfrac{1}{4}\Rightarrow \sin x = \pm \sqrt \frac{1}{4}=\pm\dfrac{1}{2}$$

Therefore, the principal solutions of the equation are:

• $latex x_1=\arcsin \left(\dfrac{1}{2}\right)=30º$
• $latex x_2=\arcsin \left(-\dfrac{1}{2}\right)=210º$

To solve this equation, the following identities are used:

$latex \sec^2=1+\tan^2x$

$latex \sec x = \dfrac{1}{\cos x}$

$latex \tan x=\dfrac{\sin x}{\cos x}$

Then, we have:

$$\cos x=\dfrac{2\tan x}{1+\tan^2x}=\dfrac{\dfrac{2\sin x}{\cos x}}{\sec^2x}=$$

$$=\dfrac{\dfrac{2\sin x}{\cos x}}{\dfrac{1}{\cos^2x}}=2\sin x \cos x$$

Therefore:

$$\cos x = 2\sin x \cos x$$

Dividing both sides by $latex \cos x$, as long as $latex \cos x\neq 0$:

$$\cos x = 2\sin x \cos x\Rightarrow \dfrac{\cos x}{\cos x}=\dfrac{2\sin x \cos x}{\cos x}$$

Leading to:

$$2\sin x=1$$

Then:

$$x=\arcsin\left(\dfrac{1}{2}\right)$$

$latex x = 30º, 150º,…-330º,… $

In general form, the solution is:

$latex x = 30º+k\cdot180º$

Or also:

$latex x=\dfrac{\pi}{6}+k\pi$

Where $latex k$ is an integer.

Using the following change of variable: $latex p = \sin x$, $latex ~q=\cos y$, the system of equations looks like this:

$$ \left\{\begin{array}{rcl} p+q&=&\sqrt{2}\\ \frac{1}{p}+\frac{1}{q}&=&2\sqrt{2} \end{array}\right.$$

From the first equation, we get:

$latex p =\sqrt{2}-q$

It is then substituted into the second equation:

$$\left(\dfrac{1}{\sqrt{2}-q}\right)+\left(\dfrac{1}{q}\right)=2\sqrt{2}$$

$$\dfrac{q+\sqrt{2}-q}{q(\sqrt{2}-q)}=2\sqrt{2}$$

$$\dfrac{\sqrt{2}}{q(\sqrt{2}-q)}=2\sqrt{2}$$

Therefore:

$$q(\sqrt{2}-q)=0.5$$

Applying the distributive property, we obtain a second-degree equation:

$latex q^2-\sqrt{2}q+0.5=0$

Whose solution is:

$latex q=\dfrac{\sqrt{2}}{2}=0.70710…$

Substituting:

$latex p =\sqrt{2}-q$

We get:

$$p =\sqrt{2}-\dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{2}}{2}=0.70710…$$

Substituting back, we get:

$$p = \sin x\Rightarrow\sin x=\dfrac{\sqrt{2}}{2}\Rightarrow x=\arcsin\left(\dfrac{\sqrt{2}}{2}\right)$$

And the general solution is:

$$x = 2k\pi+\dfrac{\pi}{4}$$

Where $latex k$ is an integer.

On the other hand:

$latex q = \cos y$

$$q = \cos y\Rightarrow\cos y=\dfrac{\sqrt{2}}{2}\Rightarrow y=\arccos\left(\dfrac{\sqrt{2}}{2}\right)$$

$$ y = 2k\pi+\dfrac{\pi}{4}$$

Where $latex k$ is an integer.