"Standard" trig equation: Now we need to move into a distinct type of trig equation. All of the trig equations solved to this point were, in some way, more or less the "standard" trig equation which is generally solved in a trig class. There are other kinds of equations involving trig functions though that we have to take a quick look at. The remaining examples illustrate some of these different kinds of trig equations.

Example   Solve 2 cos(6 y ) + 11cos (6 y ) sin (3 y ) = 0 .

Solution: Hence, definitely this doesn't look like any of the equations we've solved out to this point and initially the procedure is different as well. Firstly, notice that there is a cos(6 y ) in each term, so let's factor out that and see what we have.

                                            Cos(6 y ) (2 + 11sin (3 y )) = 0

Now we have a product of two terms which is zero and hence we know that we must have,

               Cos(6 y ) = 0       OR      2 + 11sin (3 y ) = 0

Now, at this instance we have two trig equations to solve out and each is identical to the type of equation we were solving earlier.  Due to this we won't put in much detail about solving these two equations.

Firstly, solving cos(6 y ) = 0 gives,

6 y = ?/2 + 2 ? n

                                                     y= ?/12 + ? n/3

                                                     y= ?/4 + ? n/3              n= 0, ±1, ±2,.........

6 y = 3?/2 + 2 ? n

Next, solving out 2 + 11sin (3 y) = 0 gives,

3 y = 6.1004 + 2 ? n             y= 2.0335+ 2 ? n /3         ⇒    n= 0, ±1, ±2,...........

3 y = 3.3244 + 2 ? n              y= 1.1081 + 2 ? n/3

In these notes we tend to take positive angles and hence the first solution here is in fact 2 ? - 0.1828 where our calculator provides us -0.1828 as the answer while using the inverse sine function.

The solutions to this equation are then,

y= ?/12 + ? n/3

y= ?/4 + ? n/3

y= 2.0335 + 2 ? n/3

y= 1.1081 + 2 ? n/3

 n=0, ±1, ±2,........