Solving Trigonometric Equations In Degrees - BBC Bitesize - BBC

In this guide

  1. Revise
  2. Test
  1. Solving trigonometric equations in degrees
  2. Basic trigonometric equations
  3. Further trigonometric equations
  4. Using the double angle formulae
  5. Equations involving compound angles (extension)
  6. Trigonometric equations in radians (extension)
  7. Trigonometric equations involving the wave formula

Solving trigonometric equations in degrees

Watch this video to learn about solving trigonometric equations.

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Example

Solve the equation \(\sin x^\circ = 0.5\), where \(0 \le x \textless 360\).

Solution

Let's remind ourselves of what the sine graph looks like so that we can see how many solutions we should be expecting:

Sine graph displaying two solutions

Therefore, from the graph of the function, we can see that we should be expecting 2 solutions: 1 solution being between \(0^\circ\) and \(90^\circ\) and the other solution between \(90^\circ\) and \(180^\circ\).

\(\sin x^\circ = 0.5\)

\(x^\circ = {\sin ^{ - 1}}(0.5)\)

\(x^\circ = 30^\circ\)

So we know that the first solution is \(30^\circ\) as previously predicted from the graph.

To get the other solution, we go back to our quadrants and use the appropriate rule:

Top r: 1st quadrant, x degrees, all positive. Top l: 2nd quadrant, 180-x degrees, Sin positive. Bottom l: 3rd quadrant, 180+x degrees, Tan positive. Bottom r: 4th quadrant, 360-x degrees, Cos positive

Therefore since the trig equation we are solving is sin and it is positive (0.5), then we are in the 1st and 2nd quadrants.

We have already found the first solution which is the acute angle from the 1st quadrant, so to find the second solution, we need to use the rule in the 2nd quadrant.

\(x^\circ = 180^\circ - 30^\circ\)

\(x^\circ = 150^\circ\)

\(x^\circ = 30^\circ ,\,150^\circ\)

Question

Solve the equation \(\sin x^\circ = - 0.349\), where \(0 \le x \textless 360\).

Show answer

Sine graph with two solutions when y=-0.349

From the graph of the function, we can see that we should be expecting 2 solutions: 1 solution between \(180^\circ\) and \(270^\circ\) and the other between \(270^\circ\) and \(360^\circ\).

\(\sin x^\circ = - 0.349\)

Since this is sin, but negative this means that we will be in the two quadrants where the sine function is negative - the third and fourth quadrants.

We need to firstly find the acute angle to use with the rules in these quadrants.

\(\sin x^\circ = - 0.349\)

When calculating the acute angle, we ignore the negative.

\(x^\circ = {\sin ^{ - 1}}(0.349)\)

\(x^\circ = 20.4261...\)

\(x^\circ = 20.4^\circ\) (to 1 d.p.)

Third quadrant

\(x^\circ = 180^\circ + 20.4^\circ\)

\(x^\circ = 200.4^\circ\)

Fourth quadrant

\(x^\circ = 360^\circ - 20.4^\circ\)

\(x^\circ = 339.6^\circ\)

Therefore \(x^\circ = 200.4^\circ ,\,339.6^\circ\)

Next pageBasic trigonometric equations

More guides on this topic

  • Dividing and factorising polynomial expressions
  • Laws of logarithms and exponents
  • Solving polynomial equations
  • Solving logarithmic and exponential equations
  • Trigonometric expressions
  • Identifying and sketching related functions
  • Determining composite and inverse functions

Related links

  • BBC Podcasts: Maths
  • BBC Radio 4: Maths collection
  • SQA: Higher Mathematics
  • Emaths
  • Stemnet
  • NRICH

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