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The number of all possible values of θ when θ ∈(0, π) for which the system of equation   (y + z) cos 3θ = (xyz) sin 3θ
(xyz) sin 3θ = (y + 2z)cos 3θ + y sin 3θ have a solution (x0, y0, z0) with y0,z0 ≠ 0 is
[JEE 2010]
  • a)
    0
  • b)
    3
  • c)
    4
  • d)
    2
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
The number of all possible values of θ when θ ∈(0, ...
Number of solution are 3
(y0 +z0) cos 3θ = x0y0z0 sin 3θ ....(i) y0. z0 ≠ 0
x0y0z0 sin 3θ = 2z0cos3θ + 2y0 sin 3θ ....(ii)   
x0y0z0 sin 3θ = (y0 + 2z0) cos 3θ +y0sin 3θ ....(iii)
subtracting (ii) from (iii) y0 cos 3θ = y0 sin3θ ; y0 ≠ 0 

From (i) & (ii)
(y0 - z0) cos 3θ = 2y0 sin3θ & (i) (ii)
- 2z0 cos3θ = 2y0 sin 3θ (y + z) = 0 & from (i) x0 = 0 

satisfies all the given equation
∴ number of values of θ is 3
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The number of all possible values of θ when θ ∈(0, ...
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The number of all possible values of θ when θ ∈(0, π) for which the system of equation (y + z) cos 3θ = (xyz) sin 3θ(xyz) sin 3θ = (y + 2z)cos 3θ + y sin 3θ have a solution (x0, y0, z0) with y0,z0 ≠ 0 is[JEE 2010]a)0b)3c)4d)2Correct answer is option 'C'. Can you explain this answer?
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The number of all possible values of θ when θ ∈(0, π) for which the system of equation (y + z) cos 3θ = (xyz) sin 3θ(xyz) sin 3θ = (y + 2z)cos 3θ + y sin 3θ have a solution (x0, y0, z0) with y0,z0 ≠ 0 is[JEE 2010]a)0b)3c)4d)2Correct answer is option 'C'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The number of all possible values of θ when θ ∈(0, π) for which the system of equation (y + z) cos 3θ = (xyz) sin 3θ(xyz) sin 3θ = (y + 2z)cos 3θ + y sin 3θ have a solution (x0, y0, z0) with y0,z0 ≠ 0 is[JEE 2010]a)0b)3c)4d)2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The number of all possible values of θ when θ ∈(0, π) for which the system of equation (y + z) cos 3θ = (xyz) sin 3θ(xyz) sin 3θ = (y + 2z)cos 3θ + y sin 3θ have a solution (x0, y0, z0) with y0,z0 ≠ 0 is[JEE 2010]a)0b)3c)4d)2Correct answer is option 'C'. Can you explain this answer?.
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