# What values of θ between 0 and 2π satisfy the equation cosec(θ) + 5cot(θ) = 3sin(θ)?

This is a question from a C4 paper.

The first task is to translate the equation into something you’re more familiar with: write the equation in terms of sin(θ) and cos(θ). To do this use we’ll first need the definition of cosec(θ) and cot(θ). A trick for remembering which is which is to look at the third letter:

co** s**ec(θ) = 1/

**in(θ) se**

__s__**(θ) = 1/**

__c__**os(θ) co**

__c__**(θ) = 1/**

__t__**an(θ)**

__t__So let’s employ these definitions:

cosec(θ) + 5cot(θ) = 3sin(θ)

⇒ 1/sin(θ) + 5/tan(θ) = 3sin(θ) *(using definitions of cosec(θ) and cot(θ))*

⇒ 1/sin(θ) + 5cos(θ)/sin(θ) = 3sin(θ) *(using tan(θ) = sin(θ)/cos(θ))*

⇒ 1 + 5cos(θ) = 3sin²(θ) *(multiplying both sides by sin(θ))*

Notice that we can now completely remove the sin(θ) term from the equation and leave only cos(θ) by using the identity sin²(θ) + cos²(θ) ≡ 1:

⇒ 1 + 5cos(θ) = 3(1- cos²(θ))

⇒ 3cos²(θ) + 5cos(θ) - 2 = 0

That’s much nicer. We now have a quadratic equation in terms of cos(θ).

Factorising gives (3cos(θ) - 1)(cos(θ) + 2) = 0 giving us solutions cos(θ) = 1/3 and cos(θ) = -2. Note that the second solution is not possible, since the graph of cos(θ) is bounded below by -1, so never attains the value -2. So we may discard it leaving only cos(θ) = 1/3 to worry about.

Plugging θ = arccos(1/3) into your calculator (with your calculator in radians!) gives 1.23 to 2 decimal places.

However, this is only half the answer. The question asked for solutions in the** range** **0 to 2π**. We now have to use the cos(θ) graph to find all remaining solutions. I usually do this by drawing a quick sketch of cos(θ) from 0 to 2π and a horizontal line at y = 1/3. Upon doing so you will see there are 2 solutions in this range. By the sketch and the symmetry of the cos(θ) graph you should be able to deduce the other solution is 2π-1.23 = 5.05 to 2 decimal places.

As usual, plugging the solutions you get back into the equation is a great way to check whether you’ve got the correct answer!