5.3 Equations  (Page 4/4)

Solution: $x=-320,9^{\circ };-{230}^{\circ };-140,9^{\circ };-50,9^{\circ };39,1^{\circ };129,1^{\circ };219,1^{\circ }$ and $309,1^{\circ }$

Linear trigonometric equations

Just like with regular equations without trigonometric functions, solving trigonometric equations can become a lot more complicated. You should solve these just like normal equations to isolate a single trigonometric ratio. Then you follow the strategy outlined in the previous section.

Quadratic and higher order trigonometric equations

The simplest quadratic trigonometric equation is of the form

This type of equation can be easily solved by rearranging to get a more familiar linear equation

This gives two linear trigonometric equations. The solutions to either of these equations will satisfy the original quadratic.

The next level of complexity comes when we need to solve a trinomial which contains trigonometric functions. It is much easier in this case to use temporary variables . Consider solving

Here we notice that $tan(2x+1)$ occurs twice in the equation, hence we let $y=tan(2x+1)$ and rewrite:

That should look rather more familiar. We can immediately write down the factorised form and the solutions:

Next we just substitute back for the temporary variable: $tan(2x+1)=-1\mathrm{or}tan(2\mathrm{x}+1)=-2$ And then we are left with two linear trigonometric equations. Be careful: sometimes one of the two solutions will be outside the range of the trigonometric function. In that case you need to discard that solution. For example consider the same equation with cosines instead of tangents ${cos}^2(2x+1)+3cos(2x+1)+2=0$ Using the same method we find that $cos(2x+1)=-1\mathrm{or}cos(2\mathrm{x}+1)=-2$ The second solution cannot be valid as cosine must lie between $-1$ and 1. We must, therefore, reject the second equation. Only solutions to the first equation will be valid.

More complex trigonometric equations

Here are two examples on the level of the hardest trigonometric equations you are likely to encounter. They require using everything that you have learnt in this chapter. If you can solve these, you should be able to solve anything!

Solving trigonometric equations

1. 1. Find the general solution of each of the following equations. Give answers to one decimal place.
   2. Find all solutions in the interval $\theta \in [-{180}^{\circ };{360}^{\circ }].$
      1. $sin\theta =-0,327$
      2. $cos\theta =0,231$
      3. $tan\theta =-1,375$
      4. $sin\theta =2,439$
2. 1. Find the general solution of each of the following equations. Give answers to one decimal place.
   2. Find all solutions in the interval $\theta \in [0^{\circ };{360}^{\circ }].$
      1. $cos\theta =0$
      2. $sin\theta =\frac{\sqrt{3}}{2}$
      3. $2cos\theta -\sqrt{3}=0$
      4. $tan\theta =-1$
      5. $5cos\theta =-2$
      6. $3sin\theta =-1,5$
      7. $2cos\theta +1,3=0$
      8. $0,5tan\theta +2,5=1,7$
3. 1. Write down the general solution for $x$ if $tanx=-1,12$ .
   2. Hence determine values of $x\in [-{180}^{\circ };{180}^{\circ }]$ .
4. 1. Write down the general solution for $\theta$ if $sin\theta =-0,61$ .
   2. Hence determine values of $\theta \in [0^{\circ };{720}^{\circ }]$ .
5. 1. Solve for A if $sin(\mathrm{A}+{20}^{\circ })=0,53$
   2. Write down the values of A $\in [0^{\circ };{360}^{\circ }]$
6. 1. Solve for $x$ if $cos(x+{30}^{\circ })=0,32$
   2. Write down the values of $x\in [-{180}^{\circ };{360}^{\circ }]$
7. 1. Solve for $\theta$ if ${sin}^2\left(\theta \right)+0,5sin\theta =0$
   2. Write down the values of $\theta \in [0^{\circ };{360}^{\circ }]$