A trigonometric equation is an equation that uses at least one variable inside a trigonometric function.
Here are some examples of trigonometric equations. Follow the links to see the solutions.
Note that an equation like ‘$\,5  x\sin 3 = 7\,$’ is not a trigonometric equation,
because it does not have a variable inside a trigonometric function.
Here, $\,\sin 3\,$ is just a constant.
This is a linear equation in one variable
with unique solution $\displaystyle\,x = \frac{2}{\sin 3}\,$.
The equation ‘$\,\sqrt 2\sin x  1 = 0\,$’ is a trigonometric equation because of the ‘$\,x\,$’ inside the sine function.
Note that there is no other appearance of $\,x\,$ in this equation.
In cases such as this, we start by isolating the trigonometric function containing the variable.
(Recall that isolate means to get all by itself on one side of the equation.)
$\sqrt 2\sin x  1 = 0$  original equation  
$\displaystyle\sin x = \frac{1}{\sqrt 2}$  Isolate the trigonometric function with variable input: add $\,1\,$ to both sides (using the addition property of equality); divide both sides by $\,\sqrt 2\,$ (using the multiplication property of equality). At this point, several ‘checks’ should run through your head:


reporting the solutions in degrees:
$x = 45^\circ + {360k\,}^\circ\,$ for all integers $\,k\,$, or $x = 135^\circ + {360k\,}^\circ\,$ for all integers $\,k\,$ reporting the solutions in radians: $\displaystyle x = \frac{\pi}{4} + 2\pi k\,$ for all integers $\,k\,$, or $\displaystyle x = \frac{3\pi}{4} + 2\pi k\,$ for all integers $\,k\,$ 
Precisely what angles have sine equal to $\,\frac{1}{\sqrt 2}\,$? From the special triangle below, we see that $\,\sin 45^\circ = \frac{1}{\sqrt 2}\,$; so, $\,45^\circ\,$ (or $\,\displaystyle\frac{\pi}{4}\,$ radians) is one of the solutions of the equation. Recall that the sine function gives the $y$values of points on the unit circle. From the unit circle above, we can now get all the angles with $y$value equal to $\,\frac{1}{\sqrt 2}\,$, as follows: Recall that, when laying off angles: and clockwise is the negative direction Also recall that the integers are: $$\ldots, 3,2,1,0,1,2,3,\ldots$$ Start at the terminal point for $\,45^\circ\,$, in the first quadrant:

The equation ‘$\,2\cos(3x) = 1\,$’ is a trigonometric equation because of the ‘$\,x\,$’ inside the cosine function.
Note that there is no other appearance of $\,x\,$ in this equation.
Again, we start by isolating the trigonometric function containing the variable.
$\,2\cos(3x) = 1\,$  original equation  
$\displaystyle\cos(3x) = \frac{1}{2}$  Isolate the trigonometric function with variable input. The cosine function takes on the value $\,\frac 12\,$ infinitely many times, so again there are infinitely many solutions. 

to get $\,x\,$ in degrees:
$3x = 60^\circ + {360k\,}^\circ\,$ for all integers $\,k\,$, or $3x = 60^\circ + {360k\,}^\circ\,$ for all integers $\,k\,$ to get $\,x\,$ in radians: $\displaystyle 3x = \frac{\pi}{3} + 2\pi k\,$ for all integers $\,k\,$, or $\displaystyle 3x = \frac{\pi}{3} + 2\pi k\,$ for all integers $\,k\,$ 
For the moment, ignore the ‘$\,3x\,$’ inside the cosine function. (We'll use it in a minute.) Ask: What angles have cosine equal to $\displaystyle\,\frac{1}{2}\,$? From the special triangle below, we see that $\displaystyle\,\cos 60^\circ = \frac{1}{2}\,$. Recall that the cosine function gives the $x$values of points on the unit circle. As in Example 1, get all the angles with $x$value equal to $\,\frac{1}{2}\,$: $$ 60^\circ + {360k\,}^\circ \ \ \text{ for all integers $k$ } \quad \text{(in quadrant I)} \tag{*} $$ $$ 60^\circ + {360k\,}^\circ \ \ \text{ for all integers $k$ } \quad \text{(in quadrant IV)} \tag{**} $$ These are all possible angles with cosine equal to $\,\frac 12\,$. Now, we'll use the ‘$\,3x\,$’. Note that if $\,x\,$ has units of degrees, then $\,3x\,$ also has units of degrees. The angle inside the cosine function (which in this example is $\,3x\,$) must equal one of the angles in (*) or (**), in order to have cosine equal to $\,\frac 12\,$. Thus, for all integers $\,k\,$, and for $\,x\,$ in degrees: $$ \begin{gather} 3x = 60^\circ + {360k\,}^\circ\cr \text{or}\cr 3x = 60^\circ + {360k\,}^\circ \end{gather} $$ 

degree measure: $x = 20^\circ + {120k\,}^\circ\,$ for all integers $\,k\,$, or $x = 20^\circ + {120k\,}^\circ\,$ for all integers $\,k\,$ radian measure: $\displaystyle x = \frac{\pi}{9} + \frac{2\pi k}{3}\,$ for all integers $\,k\,$, or $\displaystyle x = \frac{\pi}{9} + \frac{2\pi k}{3}\,$ for all integers $\,k\,$ 
Solve for $\,x\,$. Done! 
Graphical methods for solving equations (and inequalities) were studied thoroughly in earlier sections:
The equation ‘$\,2\cos x = 3\sin x\,$’ is a trigonometric equation because there is at least one variable inside a trigonomeric function.
In this equation, there are two trigonometric functions with variable inputs;
isolating either one would still cause a variable expression (not a constant) to appear on the other side of the equation.
Thus, the technique illustrated in Example 1 and Example 2 doesn't work.
Both sides of the equation ‘$\,2\cos x = 3\sin x\,$’ are easy to graph—it is wellsuited to graphical methods.
Graph the lefthand side; graph the righthand side; approximate the intersection point(s) in a desired interval.
$2\cos x = 3\sin x$  original equation; find solutions in the interval $\,[0,2\pi]\,$ 
$x\approx 2.6\,$ or $\,x\approx 5.7\,$ 
Graph, on the interval $\,[0,2\pi]\,$:
It is clear there are exactly two solutions in the interval $\,[0,2\pi]\,$; these correspond to the intersection points of the two graphs:
can be used to get decimal approximations for $\,x_1\,$ and $\,x_2\,$. 
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