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Find θ , if tan θ + 0 , 5 = 1 , 5 , with 0 < θ < 90 . Determine the solution graphically.

  1. tan θ + 0 , 5 = 1 , 5 tan θ = 1
  2. y = tan θ y = 1
  3. The graphs intersect at θ = 45 .

Algebraic solution

The inverse trigonometric functions arcsin , arccos and arctan can also be used to solve trigonometric equations. These may be shown as second functions on your calculator: sin - 1 , cos - 1 and tan - 1 .

Using inverse trigonometric functions, the equation

sin θ = 0 , 5

is solved as

sin θ = 0 , 5 θ = arcsin 0 , 5 = 30

Find θ , if tan θ + 0 , 5 = 1 , 5 , with 0 < θ < 90 . Determine the solution using inverse trigonometric functions.

  1. tan θ + 0 , 5 = 1 , 5 tan θ = 1 θ = arctan 1 = 45

Trigonometric equations often look very simple. Consider solving the equation sin θ = 0 , 7 . We can take the inverse sine of both sides to find that θ = sin - 1 ( 0 , 7 ) . If we put this into a calculator we find that sin - 1 ( 0 , 7 ) = 44 , 42 . This is true, however, it does not tell the whole story.

The sine graph. The dotted line represents y = 0 , 7 . There are four points of intersection on this interval, thus four solutions to sin θ = 0 , 7 .

As you can see from [link] , there are four possible angles with a sine of 0 , 7 between - 360 and 360 . If we were to extend the range of the sine graph to infinity we would in fact see that there are an infinite number of solutions to this equation! This difficulty (which is caused by the periodicity of the sine function) makes solving trigonometric equations much harder than they may seem to be.

Any problem on trigonometric equations will require two pieces of information to solve. The first is the equation itself and the second is the range in which your answers must lie. The hard part is making sure you find all of the possible answers within the range. Your calculator will always give you the smallest answer ( i.e.  the one that lies between - 90 and 90 for tangent and sine and one between 0 and 180 for cosine). Bearing this in mind we can already solve trigonometric equations within these ranges.

Find the values of x for which sin x 2 = 0 , 5 if it is given that x < 90 .

  1. Because we are told that x is an acute angle, we can simply apply an inverse trigonometric function to both sides.

    sin x 2 = 0 , 5 x 2 = arcsin 0 , 5 x 2 = 30 x = 60

We can, of course, solve trigonometric equations in any range by drawing the graph.

For what values of x does sin x = 0 , 5 , when - 360 x 360 ?

  1. We take a look at the graph of sin x = 0 , 5 on the interval [ - 360 , 360 ] . We want to know when the y value of the graph is 0 , 5 , so we draw in a line at y = 0 , 5 .

  2. Notice that this line touches the graph four times. This means that there are four solutions to the equation.

  3. Read off the x values of those intercepts from the graph as x = - 330 , - 210 , 30 and 150 .

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Source:  OpenStax, Math 1508 (lecture) readings in precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11354/1.1
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