Is Trigonometric Equations part of the EAPCET Mathematics syllabus?

Yes. Trigonometric Equations is taught in Intermediate first year · Paper IA Mathematics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.

How many Mathematics questions are there in EAPCET?

80 of the 160 questions in the Engineering paper are Mathematics, spread across 37 chapters.

What mistakes do students make in Trigonometric Equations?

Dividing both sides by sin θ or cos θ and losing the solutions where it is zero; factorise instead of cancelling. Using θ = nπ ± α for sin θ = sin α; that form belongs to sin²θ = sin²α, while sin θ = sin α gives nπ + (−1)ⁿα. Keeping roots like sin x = 2 from a factorised quadratic; reject any value outside [−1, 1]. Squaring both sides to solve and then not checking the answers, which lets extra roots slip in.

EAPCET Mathematics · Intermediate first year · Paper IA

Trigonometric Equations

Trigonometric Equations is about finding every angle that satisfies an equation — the general solution — and counting solutions within an interval.

Subject: Mathematics
Taught in: Intermediate first year · Paper IA
Mathematics in the paper: 80 of 160 questions

Key formulas and results

$\sin\theta = \sin\alpha \Rightarrow \theta = n\pi + (-1)^{n}\alpha$
$\cos\theta = \cos\alpha \Rightarrow \theta = 2n\pi \pm \alpha$
$\tan\theta = \tan\alpha \Rightarrow \theta = n\pi + \alpha$
$\sin^{2}\theta = \sin^{2}\alpha$ (and likewise $\cos^{2}$ , $\tan^{2}$ ) $\Rightarrow \theta = n\pi \pm \alpha$
$a\cos\theta + b\sin\theta = c$ has real solutions only if $|c| \le \sqrt{a^{2} + b^{2}}$
$\cos n\pi = (-1)^{n}$ and $\sin n\pi = 0$ for every integer $n$
$\cos\dfrac{n\pi}{2} = -1$ when $n = 2, 6, 10, \dots$ and $+1$ when $n$ is a multiple of 4

What this chapter covers

General solutions of sin θ = k, cos θ = k and tan θ = k
Equations reducible to quadratic form
Equations of the form a cos θ + b sin θ = c
Number of solutions in a given interval

How it is usually asked

Write the general solution of an equation
Count the solutions in [0, 2π] or another interval
Solve by factorising into simpler trigonometric equations

Understanding Trigonometric Equations

A trigonometric equation usually has infinitely many solutions because the functions repeat. The general solution captures all of them in one formula. If sin θ = sin α, then θ = nπ + (−1)ⁿα; if cos θ = cos α, then θ = 2nπ ± α; and if tan θ = tan α, then θ = nπ + α, where n is any integer. The angle α is normally chosen as the principal value.

Most equations are reduced to these basic forms. A quadratic in sin θ or cos θ is factorised, and each factor gives a basic equation, keeping only values between −1 and 1. An equation of the form a cos θ + b sin θ = c is divided by √(a² + b²) and rewritten as a single cosine, which also shows it has solutions only when |c| ≤ √(a² + b²).

EAPCET questions ask for a general solution, the number of solutions in an interval such as [0, 2π], or the solutions that satisfy an extra condition. For counting, list the solutions from the general formula for successive n, or sketch both sides as graphs and count the intersections, taking care with whether the interval endpoints are included.

Worked example

EAPCET-style question

The number of solutions of $2\sin^{2}x - 3\sin x + 1 = 0$ in $[0, 2\pi]$ is

(1)1
(2)2
(3)3
(4)4

Solution

Factorise: $(2\sin x - 1)(\sin x - 1) = 0$ .
$\sin x = \tfrac{1}{2}$ gives $x = \tfrac{\pi}{6}$ and $\tfrac{5\pi}{6}$ in $[0, 2\pi]$ .
$\sin x = 1$ gives $x = \tfrac{\pi}{2}$ only.
Total $= 2 + 1 = 3$ solutions.

Answer: 3

Shortcuts and results to remember

To count solutions, sketch the left side and the right side as two graphs on the given interval and count their points of intersection.

Common mistakes to avoid

Dividing both sides by sin θ or cos θ and losing the solutions where it is zero; factorise instead of cancelling.
Using θ = nπ ± α for sin θ = sin α; that form belongs to sin²θ = sin²α, while sin θ = sin α gives nπ + (−1)ⁿα.
Keeping roots like sin x = 2 from a factorised quadratic; reject any value outside [−1, 1].
Squaring both sides to solve and then not checking the answers, which lets extra roots slip in.

How to prepare it

After solving, check each value against the domain — values where tan or sec is undefined must be dropped.

For counting questions, sketch the graphs over the interval; it is quicker than listing every solution.

Once the formulas feel familiar, test them the way the exam does: practise Trigonometric Equations questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Mathematics chapters

See every chapter in the EAPCET Mathematics syllabus.

Practise Trigonometric Equations chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Trigonometric Equations part of the EAPCET Mathematics syllabus?: Yes. Trigonometric Equations is taught in Intermediate first year · Paper IA Mathematics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Trigonometric Equations for EAPCET?: After solving, check each value against the domain — values where tan or sec is undefined must be dropped. For counting questions, sketch the graphs over the interval; it is quicker than listing every solution.
How many Mathematics questions are there in EAPCET?: 80 of the 160 questions in the Engineering paper are Mathematics, spread across 37 chapters.
What mistakes do students make in Trigonometric Equations?: Dividing both sides by sin θ or cos θ and losing the solutions where it is zero; factorise instead of cancelling. Using θ = nπ ± α for sin θ = sin α; that form belongs to sin²θ = sin²α, while sin θ = sin α gives nπ + (−1)ⁿα. Keeping roots like sin x = 2 from a factorised quadratic; reject any value outside [−1, 1]. Squaring both sides to solve and then not checking the answers, which lets extra roots slip in.