5.3 Equations  (Page 3/4)

Calculators give us the smallest possible answer (sometimes negative) which satisfies the equation. For example, if we wish to solve $sin\theta =0,3$ we can apply the inverse sine function to both sides of the equation to find:

However, we know that this is just one of infinitely many possible answers. We get the rest of the answers by finding relationships between this small angle, $\theta$ , and answers in other quadrants. To do this we use our small angle $\theta$ as a reference angle . We then look at the sign of the trigonometric function in order to decide in which quadrants we need to work (using the CAST diagram) and add multiples of the period to each, remembering that sine, cosine and tangent are periodic (repeating) functions. To add multiples of the period we use ${360}^{\circ }·n$ (where $n$ is an integer) for sine and cosine and ${180}^{\circ }·n$ , $n\in \mathbb{Z}$ , for the tangent.

General solution using periodicity

Up until now we have only solved trigonometric equations where the argument (the bit after the function, e.g. the $\theta$ in $cos\theta$ or the $(2x-7)$ in $tan(2x-7)$ ), has been $\theta$ . If there is anything more complicated than this we need to be a little more careful. Let us try to solve $tan(2x-{10}^{\circ })=2,5$ in the range $-{360}^{\circ }\le x\le {360}^{\circ }$ . We want solutions for positive tangent so using our CAST diagram we know to look in the 1 ${}^{\mathrm{st}}$ and 3 ${}^{\mathrm{rd}}$ quadrants. Our calculator tells us that $arctan(2,5)=68,2^{\circ }$ . This is our reference angle. So to find the general solution we proceed as follows:

This is the general solution. Notice that we added the ${10}^{\circ }$ and divided by 2 only at the end. Notice that we added ${180}^{\circ }·n$ because the tangent has a period of ${180}^{\circ }$ . This is also divided by 2 in the last step to keep the equation balanced. We chose quadrants I and III because $tan$ was positive and we used the formulae $\theta$ in quadrant I and $({180}^{\circ }+\theta )$ in quadrant III. To find solutions where $-{360}^{\circ }<x<{360}^{\circ }$ we substitue integers for $n$ :

• $n=0$ ; $x=39,1^{\circ }$ ; $129,1^{\circ }$
• $n=1$ ; $x=129,1^{\circ }$ ; $219,1^{\circ }$
• $n=2$ ; $x=219,1^{\circ }$ ; $309,1^{\circ }$
• $n=3$ ; $x=309,1^{\circ }$ ; $399,1^{\circ }$ (too big!)
• $n=-1$ ; $x=-50,9^{\circ }$ ; $39,1^{\circ }$
• $n=-2$ ; $x=-140,1^{\circ }$ ; $-50,9^{\circ }$
• $n=-3$ ; $x=-230,9^{\circ }$ ; $-140,9^{\circ }$
• $n=-4$ ; $x=-320,9^{\circ }$ ; $-230,9^{\circ }$