Trigonometric Equations 1i2v16

TRIGONOMETRIC EQUATIONS

Firstly recall the graphs of y = sin x, y = cos x and y = tan x. y = sin x

Note: sin x is positive for 0º < x < 180º, and negative for 180º < x < 360º.

y = cos x

Note: cos x is positive for 0º ≤ x < 90º, and for 270º < x ≤ 360º and negative for 90º < x < 270º.

y = tan x

Note: tan x is positive for 0° < x < 90º, and for 180º < x < 270º and negative for 90º < x < 180º. and for 270º < x < 360º.

We can summarise the information in the following diagram: 90º sin +ve sin +ve cos –ve cos +ve tan –ve tan +ve º º 180 0 , 360º sin –ve sin –ve cos –ve cos +ve tan +ve tan –ve 270º This can be simplified to show just the positive ratios: 90º S sin +ve all +ve or just: 180º 0º, 360º T cos +ve tan +ve 270º

A C

N.B. You may find a mnemonic will help you memorise the positions.

Solving Trigonometric Equations CAST Rule Quadrant II

Sine       Tangent Quadrant III

Quadrant I

All     Cosine Quadrant IV

Find the measure of   0 ≤   < 3600 a) cos  = ­0.6691 b) tan  = 1.2435 The reference angle is 480.  The reference angle is 510.  The angle  is found The angle  is found in Quadrants II and III. in Quadrants I and III.

 1320 and 2280

 510 and 2310 5.2.2

Solving Trigonometric Equations with a Given Interval

Example : Solve sin x = –0.5; –360º < x ≤ 360º The sine of an angle is negative in the 3rd and 4th quadrants. α = sin–1 0.5 = 30º 



N.B. The negative is ignored to find the acute angle. Hence the solutions in the given range are: x = –150°, –30°, 210°,

330°

In problems where the angle is not simply x, the given range will need to be adjusted.

The positive ratio diagram can be used to solve trigonometric equations: Example : Using:

Solve cos x = 0.5; 0º ≤ x < 360º

S A T C

We can see that the cosine of an angle is positive in the two quadrants on the right, i.e. the 1st and 4th quadrants.  

α = cos–1 0.5 = 60º

Hence the solutions in the given range are: x = 60°, 360 – 60°

so x = 60º, 300º

Example 3: Solve 3 + 5 tan 2x = 0; 0º ≤ x ≤ 360º. Firstly we need to make tan 2x the subject of the equation: 3  tan 2x = – 5 The tangent of an angle is negative in the 2nd and 4th quadrants.  

α

tan 1

 3  °   = 30.96  5    

The range must be adjusted for the angle 2x. i.e. 0° ≤ 2x ≤ 720°. Hence: 2x = 149.04°, 329.04°, 509.04°, 689.04°. x = 74.5°, 164.5°, 254.5°, 344.5°.

Solve Trigonometric Equations • Trig equations are usually solved for a values of the variable between 0 degrees and 360 degrees. • But there are solutions outside that interval. • These other solutions differ by integral multiples of the period of the function

1 2

sin x = is a trigonometric equation. x = 6π is one of infinitely many solutions of y = sin x. y -19π 6 -3π

-11π 6

-7π 6

π 6

1

5π 6

π

-2π -π

13π 6

17π 6

2π 3π

25π 6 4π

1 x y=2

-1

All the solutions for x can be expressed in the form of a general solution. x = π + 2k π and x = 5 π + 2k π (k = 0, ±1, ± 2, ± 3,  ). 6

6

Example: Solve tan x = 1. The graph of y = 1 intersects the graph of y = tan x infinitely y many times. - π – 2π 4

-π–π 4

π 4

π + π π + 2π 4

π + 3π

4

4

y=1 -π

π

2π

3π

x

y = tan(x) x = -3π x = -π 2 2

x = π x = 3π x = 5π 2 2 2

Points of intersection are at x = π and every multiple of π added or 4 π subtracted from 4 . General solution: x = π + kπ for k any integer. 4

Example: Solve the equation 3sin x + 2 = sin x for  π ≤ x ≤ π . 2

2

y

3sin x + 2 = sin x 3sin x  sin x + 2 = 0 2sin x + 2 = 0 sin x = 

2 2

Collect like .

1

-π 1 4

x

y=-

x = 4π is the only solution in the interval  2π ≤ x ≤ 2π .

2 2

Solving Second Degree (Quadratic) Trigonometric Equations Notes

Solving Second Degree Equations Example: Solve. 0° ≤ x ≤ 360° tan2x – 3tanx – 4 = 0 let x = tan x x2 – 3x – 4 = 0 (x – 4)(x + 1) = 0 x–4=0 x=4

x+1=0 x = -1

tan x = 4

tan x = -1

x = 76° x = 256°

x = 135° x = 315°

Solving Second Degree Equations Example: Solve. 0° ≤ x ≤ 360° 3cos2x – 5cosx = 4 let x = cos x 3x2 – 5x – 4 = 0 x  b  b2  4ac x 2a

   5 

5  73 6 x  2.26 x   0.59 x

  5 2  4  3   4 2  3

Solving Second Degree Equations x = 2.26

x = -0.59

cos x = 2.26 cos x = -0.59 x=Ø x = 234°

x = 126°

Solving Second Degree Equations Example: Solve. 0° ≤ x ≤ 360° 2cos2x = cosx let x = cos x 2x2 = x 2x2 – x = 0 x(2x – 1) = 0 x = 0 2x – 1 = 0 x = .5 cos x = 0

cos x = .5

x = 90° x = 270°

x = 270°

Solving Second Degree Equations Example:

Solve. 0° ≤ x ≤ 360° 3sinx + 4 = 1/sinx let x = sin x 3x + 4 = 1/x x(3x + 4) = x(1/x) 3x2 + 4x = 1 (3x – 1)(x + 1) = 0

Solving Second Degree Equations 3x - 1= 0 x + 1 = 0 x = .333 x = -1 sin x = .333 x = 19° x = 161°

sin x = -1

x = 270°

Solving Second Degree Equations Example:

Solve. 0° ≤ x ≤ 360° 2cos2x – sinx = 1 Use the identity cos2x = 1 – sin2x 2 (1 – sin2x) – sinx = 1 2 – 2sin2x – sinx = 1 – 2sin2x – sinx + 1 = 0 2sin2x + sinx - 1 = 0

Solving Second Degree Equations 2sin2x + sinx - 1 = 0 let x = sin x 2x2 + x - 1 = 0 (2x - 1)(x + 1) = 0 2x - 1 = 0 x = ½ x = -1

x+1=0

sin x = ½ x = 30° x = 150°

sin x = -1 x = 270°

The trigonometric equation 2 sin2  + 3 sin  + 1 = 0 is quadratic in form. 2 sin2  + 3 sin  + 1 = 0 implies that (2 sin  + 1)(sin  + 1) = 0. Therefore, 2 sin  + 1 = 0 or sin  + 1 = 0. It follows that sin  = - 1 or sin  = -1. 2

Solutions:  = - π + 2kπ and  = 7π + 2kπ, from sin  = - 1 6 6 2  = -π + 2kπ, from sin  = -1

Example: Solve 8 sin  = 3 cos2  with  in the interval [0, 2π]. Rewrite the equation in of only one trigonometric function. 8 sin  = 3(1 sin2  )

Use the Pythagorean Identity.

3 sin2  + 8 sin   3 = 0.

A “quadratic” equation with sin x as the variable (3 sin   1)(sin  + 3) = 0 Factor. Therefore, 3 sin   1 = 0 or sin  + 3 = 0 Solutions: sin  = 1 or sin  = -3 3  = sin1( 1 ) = 0.3398 and  = π  sin1( 1) = 2.8107. 3 3 s

Solve: 5cos2  + cos  – 3 = 0 for 0 ≤  ≤ π. The equation is quadratic. Let y = cos  and solve 5y2 + y  3 = 0. y = (-1 ± 61 ) = 0.6810249 or -0.8810249 10 Therefore, cos  = 0.6810249 or –0.8810249. Use the calculator to find values of  in 0 ≤  ≤ π. This is the range of the inverse cosine function. The solutions are:  = cos 1(0.6810249 ) = 0.8216349 and  = cos 1(0.8810249) = 2.6488206

Example:

Solve 6 sin2 x + sin x – 1 = 0; 0º ≤ x < 360º

A quadratic equation! It may help to abbreviate sin x with s: i.e. 6s2 + s – 1 = 0 Factorising this:

(3s – 1)(2s + 1)= 0 ∴





1 sin x  3

α = 19.47º

or



1 sin x   2 

α = 30º

So, x = 19.5°, 160.5°, 210°, 330°. (To nearest 0.1º.)

Summary of key points: To solve a Trigonometric Equation: • Re-arrange the equation to make sin, cos or tan of some angle the subject. • Locate the quadrants in which the ratio is positive, or negative as required. Using:

S A T C

• Adjust the range for the given angle. • Read off all the solutions within the range. This PowerPoint produced by R. Collins; © ZigZag Education 2008–2010