SATHEE: Trigonometrical Equations 1 Question 20

Trigonometrical Equations 1 Question 20

20. The values of $θ$ lying between $θ = 0$ and $θ = π / 2$ and satisfying the equation

$| \begin{array}{ccc} 1 + \sin^{2} θ & \cos^{2} θ & 4 \sin 4 θ \\ \sin^{2} θ & 1 + \cos^{2} θ & 4 \sin 4 θ \\ \sin^{2} θ & \cos^{2} θ & 1 + 4 \sin 4 θ \end{array} | = 0, is$

(a) $7 π / 24$

(b) $5 π / 24$

(d) $π / 24$

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Answer:

Correct Answer: 20. (a, c)

Solution:

Given, $| \begin{array}{ccc} 1 + \sin^{2} θ & \cos^{2} θ & 4 \sin 4 θ \\ \sin^{2} θ & 1 + \cos^{2} θ & 4 \sin 4 θ \\ \sin^{2} θ & \cos^{2} θ & 1 + 4 \sin 4 θ \end{array} | = 0$

Applying $R_{3} \to R_{3} - R_{1}$ and $R_{2} \to R_{2} - R_{1}$ , we get

$| \begin{array}{ccc} 1 + \sin^{2} θ & \cos^{2} θ & 4 \sin 4 θ \\ - 1 & 1 & 0 \\ - 1 & 0 & 1 \end{array} | = 0$

Applying $C_{1} \to C_{1} + C_{2}$ , we get

$\begin{array}{cc} | \begin{array}{ccc} 2 & \cos^{2} θ & 4 \sin 4 θ \\ 0 & 1 & 0 \\ - 1 & 0 & 1 \end{array} | = 0 \\ \Rightarrow & 2 + 4 \sin 4 θ = 0 \\ \Rightarrow & \sin 4 θ = \frac{- 1}{2} \\ \Rightarrow & 4 θ = n π + (- 1)^{n} - \frac{π}{6} \\ \Rightarrow & θ = \frac{n π}{4} + (- 1)^{n + 1} \frac{π}{24} \end{array}$

Clearly, $θ = \frac{7 π}{24}, \frac{11 π}{24}$ are two values of $θ$ lying between 0 and $π / 2$ .

Trigonometrical Equations 1 Question 20

20. The values of $θ$ lying between $θ = 0$ and $θ = π / 2$ and satisfying the equation

Answer:

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Trigonometrical Equations 1 Question 20

20. The values of θ lying between θ=0 and θ=π/2 and satisfying the equation

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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20. The values of $θ$ lying between $θ = 0$ and $θ = π / 2$ and satisfying the equation