Quadratic Trigonometric Equations (Edexcel IGCSE Further Pure Maths): Revision Note

• Sketch the trig graphs on your exam paper

  • Then you can refer back to them as many times as you need to

• Make sure you have found all of the solutions in the given interval

  • And that you don't give solutions outside the interval 

  • For example if you get a negative solution but the interval is entirely positive

Solve the equation $11\sin x–7=5{\cos }^{2}x$, finding all solutions in the interval  $0\le x\le 2\pi$.  Give your answers correct to 3 significant figures.

${\sin }^{2}x+{\cos }^{2}x=1$  can be rearranged as  ${\cos }^{2}x=1-{\sin }^{2}x$

Substitute this to get the equation entirely in terms of $\sin x$

$11\sin x-7=5\left(1-{\sin }^{2}x\right)$

Expand the brackets and rearrange to get a quadratic equal to zero

$\begin{array}{rcl}11\sin x-7& =& 5-5{\sin }^{2}x\\ 5{\sin }^{2}x+11\sin x-12& =& 0\end{array}$

This can be solved by factorising (it might help you to think of it as  $5s^2+11s-12=0$)
You could also solve it by using the quadratic formula
Or your calculator may be able to solve quadratics

$\begin{array}{l}\left(5\sin x-4\right)\left(\sin x+3\right)=0\\ \\ \sin x=\frac{4}{5}\mathrm{or}\sin x=-3\end{array}$

$\sin x=-3$  has no solutions for $x$ because sine cannot be less than $-1$
So we only need to find solutions for $\sin x=\frac{4}{5}$

Start by finding the primary solution
The interval is given in radians, so we have to make sure the calculator is set up for radians!

$x_1={\sin }^{-1}\left(\frac{4}{5}\right)=0.927295...$

Use symmetry properties of sine to find the secondary solution

$x_2=\pi -x_1=\pi -{\sin }^{-1}\left(\frac{4}{5}\right)=2.214297...$

Both those solutions are the interval $0\le x\le 2\pi$, and there are no other solutions in the interval
(You could sketch the sine function to confirm that)

$\mathit{x}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{927}\mathbf{,}\mathbf{2}\mathbf{.}\mathbf{21}\left(3s.f.\right)$