Trigonometric equations

1 Trigonometricequationsmc-TY-trigeqn-2009 -1In this unit we consider the solution of Trigonometric equations . The strategy we adopt is to findone solution using knowledge of commonly occuring angles, and then use the symmetries in thegraphs of the Trigonometric functions to deduce additionalsolutions. Familiarity with the graphsof these functions is order to master the techniques explained here it is vital that you undertake the practiceexercises reading this text, and/or viewing the video tutorial on this topic, you should be able to: find solutions of Trigonometric equations use Trigonometric identities in the solution of Trigonometric special angles and their Trigonometric simple Trigonometric identities in the solution of examples where the interval is given in mathcentre 20091. IntroductionThis unit looks at the solution of Trigonometric order to solve these equations weshall make extensive use of the graphs of the functions sine,cosine and tangent.

2 The symmetrieswhich are apparent in these graphs, and their periodicitiesare particularly important as we Some special angles and their Trigonometric the examples which follow a number of angles and their Trigonometric ratios are used list these angles and their sines, cosines and 6 4 3 20 30 45 60 90 sin0121 2 321cos1 321 2120tan01 31 3 3. Some simple Trigonometric equationsExampleSuppose we wish to solve the equationsinx= we look for all solutions lying in theinterval0 x 360 . This means we are looking for all the angles,x, in this interval whichhave a sine of begin by sketching a graph of the functionsinxover the given interval. This is shown inFigure 1. A graph have drawn a dotted horizontal line on the graph indicating wheresinx= The solutionsof the given equation correspond to the points where this line crosses the curve. From the Tableabove we note that the first angle with a sine equal to is30.

3 This is indicated in Figure the symmetries of the graph, we can deduce all the angles which have a sine of Theseare:x= 30 ,150 This is because the second solution,150 , is the same distance to the left of180 that the firstis to the right of0 . There are no more solutions within the given mathcentre 2009 ExampleSuppose we wish to solve the equationcosx= we look for all solutions lying in theinterval0 x 360 .As before we start by looking at the graph ofcosx. This is shown in Figure 2. We have drawna dotted horizontal line wherecosx= The solutions of the equation correspond to thepoints where this line intersects the curve. One fact we do know from the Table on page 2 isthatcos 60 = + This is indicated on the graph. We can then make use of the symmetry todeduce that the first angle with a cosine equal to . This is because the angle mustbe the same distance to the right of90 that60 is to the left.

4 From the graph we see, fromconsideration of the symmetry, that the remaining solutionwe seek is240 . Thusx= 120 ,240 xxFigure 2. A graph we wish to solvesin 2x= 32for0 x 360 .Note that in this case we have the sine of a multiple angle, enable us to cope with the multiple angle we shall considera new variableuwhereu= 2x,so the problem becomes that of solvingsinu= 32for0 u 720 We draw a graph ofsinuover this interval as shown in Figure 10180o360o60o720o540o 32120o420o480osin uu-Figure 3. A graph ofsinuforulying between 0 and720 . mathcentre 2009By referring to the Table on page 2 we know thatsin 60 = 32. This is indicated on the the graph we can deduce another angle which has a sine of 32. This is120 . Because ofthe periodicity we can see there are two more angles,420 and480 . We therefore know all theangles in the interval with sine equal to 32, namelyu= 60 ,120 ,420 ,480 Butu= 2xso that2x= 60 ,120 ,420 ,480 from whichx= 30 ,60 ,210 ,240 ExampleSuppose we wish to solvetan 3x= 1for values ofxin the interval0 x 180.

5 Note that in this example we have the tangent of a multiple angle, enable us to cope with the multiple angle we shall considera new variableuwhereu= 3x,so the problem becomes that of solvingtanu= 1for0 u 540 We draw a graph oftanuover this interval as shown in Figure uuooooooooFigure 4. A graph know from the Table on page 2 that an angle whose tangent is 1is45 , so using the symmetryin the graph we can find the angles which have a tangent equal to 1. The first will be the samedistance to the right of90 that45 is to the left, that is135 . The other angles will each be180 further to the right because of the periodicity of the tangent function. Consequently thesolutions oftanu= 1are given byu= 135 ,315 ,495 ,Butu= 3xand so3x= 135 ,315 ,495 ,from whichx= 45 ,105 ,165 mathcentre 2009 ExampleSuppose we wish to solvecosx2= 12for values ofxin the interval0 x 360 .In this Example we are dealing with the cosine of a multiple angle, enable us to handle this we make a substitutionu=x2so that the equation becomescosu= 12for0 u 180 A graph ofcosuover this interval is shown in Figure uuFigure 5.

6 A graph know that the angle whose cosine is12is60 . Using the symmetry in the graph we can findall the angles with a cosine equal to 12. In the interval given there is only one angle with cosineequal to 12and that isu= 120 Butu=x2and sox= 2u. We conclude that there is a single solution,x= 240 .Let us now look at some examples over the interval 180 x 180 .ExampleSuppose we wish to solvesinx= 1for 180 x 180 .From the graph ofsinxover this interval, shown in Figure 6, we see there is only oneanglewhich has a sine equal to 1, that isx= 90 .1 1sinxx90o180o180o-90o--Figure 6. A graph of the sine mathcentre 2009 ExampleSuppose we wish to solvecos 2x=12for 180 x 180 .In this Example we have a multiple angle, handle this we letu= 2xand instead solvecosu=12for 360 x 360 A graph of the cosine function over this interval is shown in Figure uuFigure 7. A graph dotted line indicates where the cosine is equal to12.

7 Remember we already know one anglewhich has cosine equal to12and this is60 . From the graph, and making use of symmetry, wecan deduce all the other angles with cosine equal to12. These areu= 300 , 60 ,60 ,300 Thenu= 2xso that2x= 300 , 60 ,60 ,300 from whichx= 150 , 30 ,30 ,150 ExampleSuppose we wish to solvetan 2x= 3for 180 x 180 .We again have a multiple angle,2x. We handle this by lettingu= 2xso that the problembecomes that of solvingtanu= 3for 360 u 360 mathcentre 2009We plot a graph oftanubetween 360 and360 as shown in Figure 3tan uu60oooooooFigure 8. A graph know from the Table on page 2 that one angle which has a tangent equal to 3isu= 60 .We can use the symmetry of the graph to deduce others. These areu= 300 , 120 ,60 ,240 Butu= 2xand so2x= 300 , 120 ,60 ,240 and so the required solutions arex= 150 , 60 ,30 ,120 Exercise 11. Find all the solutions of each of the following equations in the given range(a)sinx=1 2for0< x <360o(b)cosx= 1 2for0< x <360o(c)tanx=1 3for0< x <360o(d)cosx= 1for0< x <360o2.

8 Find all the solutions of each of the following equations in the given range(a)tanx= 3for 180o< x <180o(b)tanx= 3for 180o< x <180o(c)cosx=12for -180o< x <180o(d)sinx= 1 2for 180o< x < mathcentre 20093. Find all the solutions of each of the following equations in the given range(a)cos 2x=1 2for 180o< x <180o(b)tan 3x= 1for 90o< x <90o(c)sin 2x=12for 180o< x <0(d)cos12x= 32for 180o< x <180o4. Using identities in the solution of equationsThere are manytrigonometric identities. Two commonly occuring ones aresin2x+ cos2x= 1sec2x= 1 + tan2xWe will now use these in the solution of Trigonometric equations . (If necessary you should referto the unit entitledTrigonometric Identities).ExampleSuppose we wish to solve the equationcos2x+ cosx= sin2xfor0 x 180 .We can use the identitysin2x+ cos2x= 1, rewriting it assin2x= 1 cos2xto write the givenequation entirely in terms of + cosx= sin2xcos2x+ cosx= 1 cos2xRearranging, we can write2 cos2x+ cosx 1 = 0 This is a quadratic equation in which the variable iscosx.

9 This can be factorised to(2 cosx 1)(cosx+ 1) = 0 Hence2 cosx 1 = 0orcosx+ 1 = 0from whichcosx=12orcosx= 1We solve each of these equations in turn. By referring to the graph ofcosxover the interval0 x 180 which is shown in Figure 9, we see that there is only one solution of the equationcosx=12in this interval, and this isx= 60 . From the same graph we can deduce the solutionofcosx= 1to bex= 180 . mathcentre 2009So there are two solutions of the original equation,60 and180 . xxFigure 9. A graph we wish to solve the equation3 tan2x= 2 sec2x+ 1for0 x 180 .In this example we will simplify the equation using the identitysec2x= 1 + tan2x= 2 sec2x+ 13 tan2x= 2(1 + tan2x) + 13 tan2x= 2 + 2 tan2x+ 1 Rearranging we can writetan2x= 3so thattanx= + 3or 3We solve each of these equations solutions oftanx= 3can be obtained by inspecting the graph in Figure 10. From theTable on page 2 we know that one angle with a tangent of 3is60.

10 There are no othersolutions in the given interval. Using the symmetry of the graph we can deduce the solution ofthe equationtanx= 3. This isx= 120 .So the given equation has two solutions,x= 60 andx= 120 .90180 3 3-60120xxtanooooFigure 10. A graph mathcentre 2009 Exercise 21. Find all the solutions of each of the following equations in the given range(a)3 cos2x 3 = sin2x 1for0< x <360o(b)5 cos2x= 4 3 sin2xfor -180o< x <180o(c)tan2x= 2 sec2x 3for -180o< x <180o(d)cos2x+ 3 cosx= sin2x 2for0< x <360o5. Some examples where the interval is given in radiansIn the previous examples, we sought solutions of equations where the angle required was measuredin degrees. We now look at some examples where the angle is measured in radians. In fact, it isadvisable to work with angles in radians because many Trigonometric equations only make sensewhen an angle is measured in this we wish to solve the equationtanx= 1for x.