2010 Paper 2 Q1 - trignometric equations

We are now on our way to the second paper. Let’s look at the first question…

Solve the equation  3cot²θ + 10cosecθ = 5  for  0° ≤ θ ≤ 360°

You will realize at the very outset, that there are two trigonometric functions,  cot and cosec. We need to put the whole equation into one function only.  To do this, we need to refer to our identities. Luckily there is the identities formula given in the formula sheet (not that you did not know this at this time).  We will use the following identity:

cot²θ = cosec²θ – 1

So, substituting that into the above equation, we get

3(cosec²θ – 1) + 10cosecθ = 5 

3 cosec²θ – 3 + 10cosecθ – 5 = 0

3 cosec²θ + 10cosecθ – 8 = 0

Now, we have a quadratic equation in cosecθ, and so we can solve it using normal quadratic functions.

3 cosec²θ + 10cosecθ – 8 = 0

(3cosecθ – 2)(cosecθ + 4) = 0

cosecθ = 2/3  or cosecθ = -4

Remembering that cosecθ = 1/sinθ, then finding the reciprocal of each of the answers to cosecθ will give us the answer for sinθ.

Therefore,

sinθ = 3/2 (inadmissible)  or  sinθ = -¼

So we only solve for sinθ = -¼

The basic angle is arc sin (0.25) = 14.5°

since sinθ is negative, then θ must lie in the 3^rd and 4^th quadrants.

Therefore,

θ = 180° + 14.5°  ,   θ = 360° - 14.5° 

   =  194.5°                     = 345.5°

θ = 194.5° , 345.5°