How to Solve Trigonometry JEE 2010 System for Exactly 3 Solutions

Q: What is the answer to the trigonometry JEE 2010 integer question?

The correct answer is 3. This counts the θ values strictly between 0 and π where the three equations possess at least one solution (x, y, z) with yz nonzero. The specific values are π/12, 5π/12 and 3π/4.

Q: How to solve the trigonometry JEE 2010 system of equations?

Introduce C = cos 3θ and S = sin 3θ. Equating equations yields yS + zC = 0. Prove C ≠ 0 (else y = 0) and S ≠ 0 (else z = 0) to satisfy yz ≠ 0. Then set y = –kC, z = kS, obtain x = 0 and y + z = 0 leading to tan 3θ = 1.

Q: Why prove C and S are nonzero in trigonometry JEE 2010?

If C = 0 then S = ±1, so yS + zC = 0 forces y = 0, violating yz ≠ 0. Similarly, S = 0 forces z = 0 when C = ±1. These contradictions must be ruled out before parametric substitution. Only then can the three valid θ values be trusted.

Q: What are common mistakes in trigonometry JEE 2010 question?

Dividing by C without first proving C ≠ 0 misses the case that forces y = 0. Incorrectly counting roots of tan 3θ = 1 inside (0, 3π) by including 13π/4 or excluding a valid one yields 4 or 2 instead of 3. Both errors produce wrong integer answers.

Trigonometry JEE 2010: 3 Valid Solutions for θ in (0, π)

JEE Advanced 2010 Mathematics Trigonometry Existence of solutions using trigonometric identities

By Founder, JEEnius - IIT Kanpur Alumni · Sep 11, 2026 · 4 min read

Hard 5 min target

The number of all possible values of $θ$ , where $0 < θ < π$ , for which the system of equations

$(y + z) \cos 3 θ = (x y z) \sin 3 θ$

$x \sin 3 θ = \frac{2 \cos 3 θ}{y} + \frac{2 \sin 3 θ}{z}$

$(x y z) \sin 3 θ = (y + 2 z) \cos 3 θ + y \sin 3 θ$

have a solution $(x_{0}, y_{0}, z_{0})$ with $y_{0} z_{0} \neq 0$ , is

Show answerAnswer

Explanation

Let

$C = \cos 3 θ$

and

$S = \sin 3 θ$

The given equations become

$(y + z) C = x y z S$

$x S = \frac{2 C}{y} + \frac{2 S}{z}$

$x y z S = (y + 2 z) C + y S$

From the first and third equations, both left sides are equal to $x y z S$ . Hence,

$(y + z) C = (y + 2 z) C + y S$

Subtracting $(y + z) C$ from both sides,

$0 = z C + y S$

So,

$y S + z C = 0$

Now check special cases.

If $C = 0$ , then the above equation gives

$y S = 0$

Since $C = 0$ , $S \neq 0$ , so

$y = 0$

This violates $y_{0} z_{0} \neq 0$ . Hence $C \neq 0$ .

If $S = 0$ , then

$z C = 0$

Since $S = 0$ , $C \neq 0$ , so

$z = 0$

This also violates $y_{0} z_{0} \neq 0$ . Hence $S \neq 0$ .

Therefore, both $S$ and $C$ are non-zero. From

$y S + z C = 0$

we can write

$y = - k C$

$z = k S$

where $k \neq 0$ .

Substitute these into the second equation:

$x S = \frac{2 C}{- k C} + \frac{2 S}{k S}$

$x S = - \frac{2}{k} + \frac{2}{k}$

$x S = 0$

Since $S \neq 0$ ,

$x = 0$

Now use the first equation:

$(y + z) C = x y z S$

Since $x = 0$ , the right side is zero, so

$(y + z) C = 0$

Since $C \neq 0$ ,

$y + z = 0$

Using $y = - k C$ and $z = k S$ ,

$- k C + k S = 0$

$S = C$

Thus,

$\sin 3 θ = \cos 3 θ$

So,

$\tan 3 θ = 1$

Hence,

$3 θ = \frac{π}{4} + n π$

Given

$0 < θ < π$

we get

$0 < 3 θ < 3 π$

So the possible values of $3 θ$ are

$\frac{π}{4}, \frac{5 π}{4}, \frac{9 π}{4}$

Therefore the possible values of $θ$ are

$\frac{π}{12}, \frac{5 π}{12}, \frac{3 π}{4}$

Hence, the number of possible values of $θ$ is

$3$

What does the trigonometry JEE 2010 integer question actually require?

The question requires you to determine the number of θ values strictly between 0 and π for which the given three equations in x, y, z possess at least one solution where the product y₀z₀ is nonzero. It is an integer-answer question from Mathematics Paper 1 carrying high difficulty. The correct count is 3.

This count emerges only after systematic elimination of invalid cases.

How do you solve the trigonometry JEE 2010 system using the official method?

Let C = cos 3θ and S = sin 3θ. Equate first and third equations to obtain (y+z)C = (y+2z)C + yS, which simplifies to yS + zC = 0.

If C = 0, then yS = 0. Since C = 0 implies S = ±1, this forces y = 0. This contradicts y₀z₀ ≠ 0, so C ≠ 0. If S = 0, then zC = 0. Since S = 0 implies C = ±1, this forces z = 0, again contradicting y₀z₀ ≠ 0. Hence S ≠ 0.

Both C and S are therefore nonzero. From yS + zC = 0, set y = –kC and z = kS where k ≠ 0. Substitute these into the second equation to obtain xS = 0, hence x = 0.

With x = 0 the first equation reduces to (y + z)C = 0, so y + z = 0. Substituting the parametric forms yields S = C or tan 3θ = 1.

Solve 3θ = π/4 + nπ for integer n with 0 < 3θ < 3π to get 3θ = π/4, 5π/4, 9π/4 and therefore θ = π/12, 5π/12, 3π/4.

The number of such θ is exactly 3.

What algebraic slip turns the correct count of 3 into 2 or 4?

Treating the relation yS + zC = 0 by dividing by C without first proving C ≠ 0 hides the contradiction when C = 0 that forces y = 0 and violates the nonzero product condition.

Solving tan 3θ = 1 but incorrectly counting the roots in (0, 3π) by including 13π/4 or forgetting the open upper bound 3π leads to answer 4 or 2. Either mistake produces an incorrect single-digit answer instead of 3.

Why must you separately prove C ≠ 0 and S ≠ 0 before substituting?

The equation yS + zC = 0 links y and z directly to the trig functions. When C = 0 the equation forces y = 0 (S = ±1), violating y₀z₀ ≠ 0. The symmetric argument holds for S = 0: the equation reduces to zC = 0 and C = ±1 forces z = 0, again contradicting the given condition.

Only after discarding these cases can the substitution y = –kC, z = kS proceed safely. The final valid θ values π/12, 5π/12, 3π/4 survive solely because the special-case analysis removes the branches that break the nonzero product rule. This elimination is non-negotiable.

Which other hard trigonometry problems test the same existence and counting skills?

Find the number of θ ∈ (0, π) such that the equations x cos θ + y sin θ = 1, x sin θ – y cos θ = 0 admit nonzero (x, y) solutions.

Determine how many real solutions (x, y, z) exist for the system involving sin 2α and cos 2α with the condition xyz ≠ 0 (adapted from another Advanced year).

A third variant asks to solve tan 3θ = 1 together with the auxiliary linear relation 2 sin 3θ – cos 3θ = 0 and count roots inside (0, π).

These problems reward the same discipline: derive linear relations between variables, discard forbidden zero cases, then count roots inside the exact interval.

What repeatable checklist works for any JEE Advanced trig-system existence question?

Introduce C and S immediately and never divide before proving they are nonzero. Eliminate one variable or obtain a linear relation between y and z first. Always verify the interval 0 < θ < π translates to 0 < 3θ < 3π and list every root of the resulting equation. Confirm that the obtained (x, y, z) indeed satisfies all three original equations.

Search the past-paper archive for every similar integer question from 2007 onward and solve them under timed conditions.

How should Class-12 students and droppers handle this difficulty tier?

This question is tagged difficulty 4 and expected to take 5 minutes; practice the special-case check until it is automatic. For another 2010 problem that mixes trig with algebra, see the Mixed Mathematics JEE 2010 Matrix Match Solution. Losing 1 mark here because of an extra or missed root directly affects the Advanced rank.

When a fresh variant blocks you, photograph a doubt for a step-by-step solution with all cases shown. That turns this style of problem into reliable marks.

Frequently asked questions

What is the answer to the trigonometry JEE 2010 integer question?

The correct answer is 3. This counts the θ values strictly between 0 and π where the three equations possess at least one solution (x, y, z) with yz nonzero. The specific values are π/12, 5π/12 and 3π/4.

How to solve the trigonometry JEE 2010 system of equations?

Introduce C = cos 3θ and S = sin 3θ. Equating equations yields yS + zC = 0. Prove C ≠ 0 (else y = 0) and S ≠ 0 (else z = 0) to satisfy yz ≠ 0. Then set y = –kC, z = kS, obtain x = 0 and y + z = 0 leading to tan 3θ = 1.

Why prove C and S are nonzero in trigonometry JEE 2010?

If C = 0 then S = ±1, so yS + zC = 0 forces y = 0, violating yz ≠ 0. Similarly, S = 0 forces z = 0 when C = ±1. These contradictions must be ruled out before parametric substitution. Only then can the three valid θ values be trusted.

What are common mistakes in trigonometry JEE 2010 question?

Dividing by C without first proving C ≠ 0 misses the case that forces y = 0. Incorrectly counting roots of tan 3θ = 1 inside (0, 3π) by including 13π/4 or excluding a valid one yields 4 or 2 instead of 3. Both errors produce wrong integer answers.

integer questionjee 2010jee advancedsystem of equationstrig equationstrigonometry

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