Some equations which involve trigonometric functions of the unknown may be readily solved by using simple algebraic ideas (as Equation 1 below), while others may be impossible to solve exactly, only approximately (e.g., Equation 2 below):

EXAMPLE 1: Find all solutions of the equation
.
Solution: We can graphically visualize all the angles u which satisfy the equation by noticing that
is the y-coordinate of the point where the terminal side of the angle u (in standard position) intersects the unit circle (see Figure 1):
We can see that there are two angles in
that satisfy the equation:
and
. Since the period of the sine function is
, it follows that all solutions of the original equation are:

Find all solutions of the equation
.
Solution: Let u=2A; the equation is then equivalent to
, for which the solutions are (see EXAMPLE 1):

Hence the solutions for A are:

Find all solutions of the equation
that lie in the interval 
Solution: The left hand side of the equation can be factored as:

hence either
or
. For
,

while

and

The solution set of the original equation is then

Solve the equation
. Restrict solutions to the interval
.
Solution: The substitution
yields the equation
, which is quadratic in u. We use the quadratic formula to solve for
:

If
the calculator gives x=1.1191 as the acute solution, so we deduce that the other solution is
. The equation
has no solutions, since
for all x. The solution set is therefore
.
EXAMPLE 1: Find all solutions of the equation
Solution: We can graphically visualize all the angles u which satisfy the equation by noticing that
We can see that there are two angles in
Find all solutions of the equation
Solution: Let u=2A; the equation is then equivalent to
Hence the solutions for A are:
Find all solutions of the equation
Solution: The left hand side of the equation can be factored as:
hence either
while
and
The solution set of the original equation is then
Solve the equation
Solution: The substitution
If
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