# Tips for Solving __Conditional__ Trigonometric
Equations

Trigonometric equations fall into one of two categories:
__identities__ where the equation is true for all values of the

variable (we “establish” trigonometric identities), and c__onditional equations
__where the equation is true for only

certain values of the variable (we “solve” conditional trigonometric equations).

**Algebraic Techniques Used to Simplify Conditional
Trigonometric Equations**

Our goal is to use standard algebraic techniques to simplify an equation until
it reaches one of two forms:

1. a linear equation consisting of a __single__ trigonometric function equal
to a constant, or

2. a f__actorable__ equation consisting of one or more trigonometric
functions set equal to zero.

Algebraic techniques that can be used to simplify a
trigonometric equation into one of these two forms include:

· Add/subtract the same constant or variable term from both sides of the
equation.

· Multiply/divide both sides of the equation by the same NON–ZERO constant or
variable term.

o Note: When multiplying or dividing by a variable term you will need to put on
a restriction that the

term is not equal to zero.

o Note: Also, if the original equation contains tanθ ,
cscθ , sacθ , or cotθ
then you will also need to

put on a restriction that specifies that the denominator is not equal to zero.

· Take the square root of both sides of an equation and be sure to include the
“±” symbol.

**Working with Factorable Trigonometric Equations**

If the trigonometric equation appears to be something other than a linear
equation consisting of a single

trigonometric function, then it should be factorable once we get it in the right
form. Try to get the trigonometric

equation to look like a polynomial in a s__ingle variable__, such as all cosθ
, then check to see if you have a …

· Quadratic binomial where you just need to factor out the GCF.

· Quadratic binomial that factors as a Difference of Perfect Squares.

· Quadratic trinomial that factors into the product of two binomials using
standard factoring techniques.

In order to get your trigonometric equation consisting of a single variable term
you might need to use a formula and

simplify. If the trigonometric equation has …

· Two or more trigonometric functions and one of them is quadratic, try using
one of the Pythagorean

Identities, then simplify to get a factorable equation.

· A sum or difference of terms, and involves sinθ
and/or cos θ , try a Sum–to–Product formula to get a

factorable product.

· Two or more different functions involving two different angles such as …

o θ and 2θ , keep the
θ then use a Double Angle Formula on the 2θ
.

o 2θ and 4θ , keep the 2θ
then use a Double Angle Formula on 4θ = 2[2θ
].

· An angle of – θ , use your Even–Odd Identities to
replace it with a trigonometric function in θ .

· When all else fails, take your equation to sines and cosines and then multiply
through to clear your

denominator(s), if any. Be sure to state any restrictions.

**Writing your Solutions**

The amount of time it takes to format your solution depends on whether the
argument of the trigonometric function

is θ or something other than θ
.

[0, 2π ) | All Solutions | |

θ | List off all the solutions that lie in [0, 2π ). | List off all the solutions that lie in [0, 2π ), then after each one put “+ 2πk” where k is any integer. |

Something other than θ |
Take the argument and set it equal to each
solution that lies in [0, 2π ), then after each one put “+ 2πk” where k is any integer. Next, solve for θ in the resulting equation(s). To generate only those solutions that lie in [0, 2π ), let k = –1, k = 0, k = 1, k = 2 and so on until you have some θ < 0 (which reject), 0≤ θ < 2π (accept these as your answers), and θ ≥ 2π (which reject). |
Take the argument and set it equal to each solution that lies in [0, 2π ) then after each one put “+ 2πk” where k is any integer. Lastly, solve for θ in the resulting equation(s). |

Finally, apply your restrictions, and remember it’s okay to have “No Solution” as an answer.