Photo AI

Last Updated Sep 27, 2025

Linear Trigonometric Equations Simplified Revision Notes

Revision notes with simplified explanations to understand Linear Trigonometric Equations quickly and effectively.

user avatar
user avatar
user avatar
user avatar
user avatar

475+ students studying

5.3.2 Linear Trigonometric Equations

Linear trigonometric equations are equations that involve trigonometric functions like sine, cosine, or tangent, and the variable (usually x or θ \ x \ or \ \theta\ ) appears in a linear form, meaning it's not squared, cubed, etc. Solving these equations involves finding all possible angles that satisfy the equation within a given range, often within one full cycle of the trigonometric function (e.g., 0 \ 0^\circ) to (360 or 0 to 2π radians).(360^\circ\ or \ 0\ to \ 2 \pi \ radians).

1. Basic Steps to Solve Linear Trigonometric Equations:

  1. Isolate the Trigonometric Function:
  • Get the trigonometric function like (sinθ), (cosθ),or (tanθ)\ (\sin \theta), \ (\cos \theta ), or \ ( \tan \theta) on one side of the equation.
  1. Solve for the Angle:
  • Use the inverse trigonometric function to find the principal solution. Remember that trigonometric functions are periodic, so there may be multiple solutions within the given interval.
  1. Consider the General Solution:
  • For sine and cosine functions, account for their periodic nature by adding the general solution using θ+360norθ+2π\theta + 360^\circ n or \theta + 2\pi nn (where n \ n is an integer) to find all solutions within the specified interval.
  • For tangent, since its period is  180 or π radians, use θ +180 n or θ+π\ 180^\circ\ or \ \pi \ radians,\ use \ \theta \ + 180^\circ \ n \ or \ \theta + \pi nn to find all solutions.
  1. Check the Interval:
  • Make sure that the solutions you find are within the given interval (e.g.,  0 to 360\ 0^\circ\ to \ 360^\circ).

2. Examples of Solving Linear Trigonometric Equations:

infoNote

Example 1: Solving sinθ\sin \theta = 0.5


  • Step 1: Isolate the function: sinθ=0.5\sin \theta = 0.5

  • Step 2: Find the principal solution: θ=sin1(0.5)\theta = \sin^{-1}(0.5) The principal solution is θ=30(sincesin30=0.5).\theta = 30^\circ (since \sin 30^\circ = 0.5).

  • Step 3: Consider all solutions: θ=30orθ=18030=150\theta = 30^\circ \quad \text{or} \quad \theta = 180^\circ - 30^\circ = 150^\circ (Because sin\sin is positive in the first and second quadrants.)

  • Step 4: List all solutions within the interval  0 to 360:\ 0^\circ \ to \ 360^\circ: θ=30,150\theta = 30^\circ, 150^\circ
infoNote

Example 2: Solving cosθ=12\cos \theta = -\frac{1}{2}


  • Step 1: Isolate the function: cosθ=12\cos \theta = -\frac{1}{2}

  • Step 2: Find the principal solution: θ=cos1(12)\theta = \cos^{-1}\left(-\frac{1}{2}\right) The principal solution is θ=120(sincecos120=12).\theta = 120^\circ (since \cos 120^\circ = -\frac{1}{2}).

  • Step 3: Consider all solutions: θ=120orθ=360120=240\theta = 120^\circ \quad \text{or} \quad \theta = 360^\circ - 120^\circ = 240^\circ (Because  cos\ \cos is negative in the second and third quadrants.)

  • Step 4: List all solutions within the interval 0 to 360: \ 0^\circ\ to \ 360^\circ: θ=120,240\theta = 120^\circ, 240^\circ
infoNote

Example 3: Solving 2sinθ+1=0 2\sin \theta + 1 = 0


  • Step 1: Isolate the function: 2sinθ=1sinθ=122\sin \theta = -1 \quad \Rightarrow \quad \sin \theta = -\frac{1}{2}

  • Step 2: Find the principal solution: θ=sin1(12)\theta = \sin^{-1}\left(-\frac{1}{2}\right) The principal solution is θ=30,but to stay within the 0 to 360\theta = -30^\circ, but \ to \ stay\ within\ the \ 0^\circ\ to \ 360^\circrange, we convert this to 330(since sin(36030=12.) \ 330^\circ (since \ \sin(360^\circ - 30^\circ = -\frac{1}{2}.)

  • Step 3: Consider all solutions: θ=180+30=210andθ=330\theta = 180^\circ + 30^\circ = 210^\circ \quad \text{and} \quad \theta = 330^\circ (Because sin\sin is negative in the third and fourth quadrants.)

  • Step 4: List all solutions within the interval  0 to 360:\ 0^\circ\ to \ 360^\circ: θ=210,330\theta = 210^\circ, 330^\circ
infoNote

Example 4: Solving tanθ=1\tan \theta = 1


  • Step 1: Isolate the function: tanθ=1\tan \theta = 1

  • Step 2: Find the principal solution: θ=tan1(1)\theta = \tan^{-1}(1) The principal solution is θ=45(sincetan45=1).\theta = 45^\circ (since \tan 45^\circ = 1).

  • Step 3: Consider all solutions: θ=45+180n\theta = 45^\circ + 180^\circ n (Because the period of  tan is 180.)\ \tan\ is \ 180^\circ.)

  • Step 4: List all solutions within the interval  0 to 360:\ 0^\circ\ to \ 360^\circ: θ=45,225\theta = 45^\circ, 225^\circ

3. General Solutions for Trigonometric Equations:

For any trigonometric equation, the general solutions are often given as:

  • Sine Equation sinθ=k:\sin \theta = k: θ=sin1(k)+360norθ=180sin1(k)+360n\theta = \sin^{-1}(k) + 360^\circ n \quad \text{or} \quad \theta = 180^\circ - \sin^{-1}(k) + 360^\circ n
  • Cosine Equation cosθ=k:\cos \theta = k: θ=cos1(k)+360norθ=cos1(k)+360n\theta = \cos^{-1}(k) + 360^\circ n \quad \text{or} \quad \theta = -\cos^{-1}(k) + 360^\circ n
  • Tangent Equation tanθ=k:\tan \theta = k: θ=tan1(k)+180n\theta = \tan^{-1}(k) + 180^\circ n

Summary:

Linear trigonometric equations involve solving for angles that satisfy trigonometric equations like sinθ=k,cosθ=k,ortanθ=k.\sin \theta = k, \cos \theta = k, or \tan \theta = k. By isolating the trigonometric function, finding the principal solution, and considering the periodic nature of trigonometric functions, you can find all solutions within a specified interval.


Solving Trig Equations

infoNote

Example: Solve sinx=0.15\sin x = -0.15 for 0x4500 \leq x \leq 450

x=sin1(0.15)=8.6269x = \sin^{-1}(-0.15) = -8.6269^\circ

Visual representation of sinx=0.15\sin x = -0.15 on the graph with solutions indicated.

x=180+8.6269,360+8.6269x = 180 + 8.6269, \quad 360 + 8.6269Solutions: 189,351\text{Solutions: } 189^\circ, 351^\circ

infoNote

Solving cos(2x)=0.2\cos(2x) = 0.2 for 0x3600 \leq x \leq 360

02x720Double limits since angle is 2x, not x0 \leq 2x \leq 720 \quad \Rightarrow \quad \text{Double limits since angle is } 2x, \text{ not } x

cosx=0.1(÷2\xcancel{\cos x=0.1 (\div 2} The only think which can penetrate is cos bracket is cos1cos^{-1}

Method:

  1. Cosine both sides:
cos1(cos(2x))=cos1(0.2)\cos^{-1}(\cos(2x)) = \cos^{-1}(0.2) 2x=cos1(0.2)2x = \cos^{-1}(0.2) 2x=78.4632x = 78.463^\circ
  1. Do not ÷2\div 2 YET!!! There are multiple solutions at the moment we use an inverse trig function. Find these.
2x=78.463,281.537,438.463,641.5372x = 78.463^\circ, 281.537^\circ, 438.463^\circ, 641.537^\circ image
  1. Once we have multiple solutions, use them to find x:
x=39.2,141,219,321x = 39.2^\circ, 141^\circ, 219^\circ, 321^\circ
infoNote

Example

Solve tan(3x)=0.27\tan(3x) = 0.27 for 180x180-180 \leq x \leq 180

  1. Rearrange the equation to solve for 3x3x:
3x=tan1(0.27)=15.10963x = \tan^{-1}(0.27) = 15.1096^\circ

Extend the interval to accommodate the factor of 33:

5403x540-540 \leq 3x \leq 540
  1. List out the angles:
3x=524.890,344.890,164.890,15.1096,195.1096,375.1096,3x = -524.890, -344.890, -164.890, 15.1096, 195.1096, 375.1096,
  1. Divide by 33 to find xx:
x=175,115,55.0,5.04,65.0,125x = -175, -115, -55.0, 5.04, 65.0, 125

infoNote

Solve sin(2x+15)=22\sin(2x + 15) = \frac{\sqrt{2}}{2} for 0x1800 \leq x \leq 180

  1. Set up the inequality:
152x+1537515 \leq 2x + 15 \leq 375

Find the related angle by taking the inverse sine:

2x+15=sin1(22)=452x + 15 = \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = 45^\circ image
  1. Use the graph to identify additional solutions: Graph shows angles at 135° and 405°.
2x+15=45,135,4052x + 15 = 45, 135, 405
  1. Solve for xx:
x=15,60x = 15, 60

Trigonometric Equations

Graphs:

  1. y=sin(x)y = \sin(x) - Red graph

    image
  2. y=cos(x)y = \cos(x) - Blue graph

    image
  3. y=tan(x)y = \tan(x) - Green graph

    image

Note: Each graph is labeled with angles in degrees on the x-axis and the function values on the y-axis, showing typical periodic behavior.


Description:

Trigonometric equations, due to the infinite nature of the standard trig functions, typically have an infinite number of solutions. For this reason, in any given equation, we are told the domain in which the solutions are required.

infoNote

Example:

Solvecos(x)=0.20x360.This is the domain in which solutions are requiredSolve \quad \cos(x) = 0.2 \\ 0^\circ \leq x \leq 360^\circ. \quad \text {This is the domain in which solutions are required}
  1. Find the first solution on your calculator:
arccos(cos(x))=arccos(0.2)\arccos(\cos(x)) = \arccos(0.2)

Note: This is not required as a line of working, but it's part of the thought process.

x=arccos(0.2)    x=78.46x = \arccos(0.2) \implies x = 78.46^\circ

Implied truncation rather than rounding.

  • The calculation is done using a calculator, displaying cos1(0.2)=78.46304097\cos^{-1}(0.2) = 78.46304097.

Steps to Solve Trigonometric Equations:

  1. Sketch the relevant graph within the domain given to identify other solutions.

  2. Conclude with all valid solutions:

  • x=78.46,281.5(4sf)\boxed {x = 78.46^\circ, 281.5^\circ \quad \text{(4sf)}}

infoNote

Example:

Solve tan(x)=0.3\tan(x) = -0.3, for 0x5400 \leq x \leq 540:

  • Initial solution: x=arctan(0.3)=16.699x = \arctan(-0.3) = -16.699^\circ

  • Note: The initial solution is invalid, but if the graph is extended, it can be used to find valid solutions.

  • Mark the graph and use extensions to find solutions:

  • x=163.3,343.3,523.3x = 163.3^\circ, 343.3^\circ, 523.3^\circ


Solving Trig Equations Involving Compound Angles:

By "compound angle," we mean "an angle more complicated than just θ\theta."

  • Example: Solve sin(3θ)=0.42,0θ180.\sin(3\theta) = 0.42, 0 \leq \theta \leq 180.
  • Notice we are solving for θ\theta, but the angle is 3θ3\theta within the sin\sin function.
  • Method: 3. Modify the domain to find limits for the compound angle in the bracket. 4. Here,03θ540. 0 \leq 3\theta \leq 540.
  • Solve within the modified domain.
Books

Only available for registered users.

Sign up now to view the full note, or log in if you already have an account!

500K+ Students Use These Powerful Tools to Master Linear Trigonometric Equations

Enhance your understanding with flashcards, quizzes, and exams—designed to help you grasp key concepts, reinforce learning, and master any topic with confidence!

50 flashcards

Flashcards on Linear Trigonometric Equations

Revise key concepts with interactive flashcards.

Try Maths Pure Flashcards

5 quizzes

Quizzes on Linear Trigonometric Equations

Test your knowledge with fun and engaging quizzes.

Try Maths Pure Quizzes

29 questions

Exam questions on Linear Trigonometric Equations

Boost your confidence with real exam questions.

Try Maths Pure Questions

27 exams created

Exam Builder on Linear Trigonometric Equations

Create custom exams across topics for better practice!

Try Maths Pure exam builder

12 papers

Past Papers on Linear Trigonometric Equations

Practice past papers to reinforce exam experience.

Try Maths Pure Past Papers

Other Revision Notes related to Linear Trigonometric Equations you should explore

Discover More Revision Notes Related to Linear Trigonometric Equations to Deepen Your Understanding and Improve Your Mastery

96%

114 rated

Trigonometric Equations

Trigonometry - Simple Identities

user avatar
user avatar
user avatar
user avatar
user avatar

259+ studying

186KViews

96%

114 rated

Trigonometric Equations

Quadratic Trigonometric Equations

user avatar
user avatar
user avatar
user avatar
user avatar

325+ studying

184KViews

96%

114 rated

Trigonometric Equations

Strategy for Trigonometric Equations

user avatar
user avatar
user avatar
user avatar
user avatar

485+ studying

199KViews
Load more notes

Join 500,000+ A-Level students using SimpleStudy...

Join Thousands of A-Level Students Using SimpleStudy to Learn Smarter, Stay Organized, and Boost Their Grades with Confidence!

97% of Students

Report Improved Results

98% of Students

Recommend to friends

500,000+

Students Supported

50 Million+

Questions answered