Specific Solution of Equation MCQ Quiz in বাংলা - Objective Question with Answer for Specific Solution of Equation - বিনামূল্যে ডাউনলোড করুন [PDF]

CONCEPT:

The general solution of the equation tan x = tan α is given by: x = nπ + α, where  and n ∈ Z

Note: The solutions of a trigonometric equation for which 0 ≤ x

CALCULATION:

Given: 

As we know that, 

⇒ 

As we know that, if tan x = tan α then x = nπ + α, where  and n ∈ Z

⇒ x = nπ + (5π/6) where n ∈ Z.

As we know that, if the solutions of a trigonometric equation for which 0 ≤ x

So, the principal solutions of the given equation are x = 5π/6 and 11π/6

Hence, the correct option is 3.

Concept:

If tan θ = tan α then θ = nπ + α, α ∈ (-π/2, π/2), n ∈ Z.

Calculation:

Given: tan x = 1/√3 such that x ∈ [0, π /2]

As we know that, tan (π/6) = 1/√3

⇒ tan x = 1/√3 = tan (π/6)

As we know that, If tan θ = tan α then θ = nπ + α, α ∈ (-π/2, π/2), n ∈ Z.

⇒ x = nπ + π/6, where n ∈ Z.

For n = 0, x = π/6 ∈ [0, π /2]

So, x = π/6 is a solution of the given equation.

For n ≥ 1, x = nπ + π/6 ∉ [0, π /2]

For n Hence, x = π/6 is the only solution of the given equation

Calculation: 

⇒ 

⇒ 

⇒ 

⇒ 

∴ Number of solution = 5  

Hence, the correct answer is Option 5. 

Calculation:

Given: sin 2x + 4 cos x = 2 + sin x

⇒ 2 sin x cos x + 4 cos x = 2 + sin x

⇒ 2 sin x cos x − sin x + 4 cos x − 2 = 0

Group terms: sin x(2 cos x − 1) + 4 cos x − 2 = 0

Let f(x) = sin x(2 cos x − 1) + 4 cos x − 2

Solve f(x) = 0 in [−π, 4π]

Use graphical methods or trial substitution over each period

Check critical intervals:

• In each period of 2π, this equation yields 1 solution.
• Total interval length = 4π − (−π) = 5π
• Hence, it spans 2 full periods of 2π + 1π = 2 solutions + additional interval
• Final solution count = 4

∴ Total number of solutions in [−π, 4π] is 4

Given

In the interval ,

So number of solutions is

We know that, .

So, from , we get that .

So, we have to find the solutions of the equation , where .

From this data, we get that or .

That is, or .

Given equation is ,

The number of common solutions in the given interval are .

Let

where is a positive integer

Now,

for and for

and

Hence, the required principal solutions are and .