SATHEE: Trigonometrical Equations 3 Question 1

Trigonometrical Equations 3 Question 1

1. For $x \in (0, π)$ , the equation $\sin x + 2 \sin 2 x - \sin 3 x = 3$ has

(2014 Adv.)

(a) infinitely many solutions

(b) three solutions

(d) no solution

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Answer:

Correct Answer: 1. (d)

Solution:

PLAN For solving this type of questions, obtain the LHS and RHS in equation and examine, the two are equal or not for a given interval.

Given, trigonometrical equation

$(\sin x - \sin 3 x) + 2 \sin 2 x = 3$

$\Rightarrow - 2 \cos 2 x \sin x + 4 \sin x \cos x = 3$

$\begin{aligned} [∵ \sin C - \sin D = 2 \cos \frac{C + D}{2} \sin \frac{C - D}{2} and \\ \sin 2 θ = 2 \sin θ \cos θ] \\ \Rightarrow 2 \sin x (2 \cos x - \cos 2 x) = 3 \\ \Rightarrow 2 \sin x (2 \cos x - 2 \cos^{2} x + 1) = 3 \\ \Rightarrow 2 \sin x \frac{3}{2} - 2 \cos x - \frac{1}{2} = 3 \\ \Rightarrow 3 \sin x - 3 = 4 \cos x - {\frac{1}{2}}^{2} \sin x \end{aligned}$

As $x \in (0, π)$ LHS $\leq 0$ and RHS $\geq 0$

For solution to exist, LHS $=$ RHS $= 0$

Now, LHS $= 0$

$\begin{aligned} \Rightarrow & 3 \sin x - 3 & = 0 \\ \Rightarrow & \sin x & = 1 \\ \Rightarrow & x & = \frac{π}{2} \end{aligned}$

For $x = \frac{π}{2}$ ,

RHS $= 4 \cos \frac{π}{2} - {\frac{1}{2}}^{2} \sin \frac{π}{2} = 4 \frac{1}{4} (1) = 1 \neq 0$

$∴$ No solution of the equation exists.

Trigonometrical Equations 3 Question 1

1. For $x \in (0, π)$ , the equation $\sin x + 2 \sin 2 x - \sin 3 x = 3$ has

Answer:

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Trigonometrical Equations 3 Question 1

1. For x∈(0,π), the equation sin⁡x+2sin⁡2x−sin⁡3x=3 has

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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1. For $x \in (0, π)$ , the equation $\sin x + 2 \sin 2 x - \sin 3 x = 3$ has