The solutions of the equation- 1-sin2x sin2x sin2x- cos2x 1-cos2x cos2x- 4sin2x 4sin2x 1-4sin2x - 0 -0 are-

Explanation for the correct optionR_1→R_1+R_20ex0ex|[ 2 2 1; cos^2x 1+cos^2x cos^2x; 4sin2x 4sin2x 1+4sin2x ]|=00ex0exC_1→C_1 C_20e ...

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Standard XII

Mathematics

Algebra of Derivatives

The solutions...

Question

The solutions of the equation $|\begin{matrix} 1 + \sin^{2} x & \sin^{2} x & \sin^{2} x \\ \cos^{2} x & 1 + \cos^{2} x & \cos^{2} x \\ 4 \sin 2 x & 4 \sin 2 x & 1 + 4 \sin 2 x \end{matrix}| = 0, (0 < x < π)$ are:

$\frac{π}{6}, \frac{5 π}{6}$

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$\frac{7 π}{12}, \frac{11 π}{12}$

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$\frac{5 π}{12}, \frac{7 π}{12}$

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$\frac{π}{12}, \frac{π}{6}$

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Solution

The correct option is B
$\frac{7 π}{12}, \frac{11 π}{12}$

Explanation for the correct option
$R_{1} \to R_{1} + R_{2} |\begin{matrix} 2 & 2 & 1 \\ \cos^{2} x & 1 + \cos^{2} x & \cos^{2} x \\ 4 \sin 2 x & 4 \sin 2 x & 1 + 4 \sin 2 x \end{matrix}| = 0 C_{1} \to C_{1} - C_{2} |\begin{matrix} 0 & 2 & 1 \\ - 1 & 1 + \cos^{2} x & \cos^{2} x \\ 0 & 4 \sin 2 x & 1 + 4 \sin 2 x \end{matrix}| = 0 ∴ 2 + 8 \sin 2 x - 4 \sin 2 x = 0 \Rightarrow \sin 2 x = - \frac{1}{2} \Rightarrow x = \frac{7 π}{12}, \frac{11 π}{12}$
Hence the correct option is option(b) i.e. $\frac{7 π}{12}, \frac{11 π}{12}$

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Similar questions

The maximum value of $f (x) = |\begin{matrix} \sin^{2} x & 1 + \cos^{2} x & \cos 2 x \\ 1 + \sin^{2} x & \cos^{2} x & \cos 2 x \\ \sin^{2} x & \cos^{2} x & \sin 2 x \end{matrix}|$ $, x \in R$ is:

Solution of the equation $x |x + 1| + 1 = 0$

Q. If

z_{1}, z_{2}, z_{3}

and

z_{4}

are the roots of the equation

z^{4} + z^{3} + z^{2} + z + 1 = 0

, then

\begin{matrix} Column I & Column II (A) & ∣ ∣ ∣ ∣ 4 \sum i = 1 (z_{i})^{4} ∣ ∣ ∣ ∣ is equal to & (p) & 0 (B) & 4 \sum i = 1 (z_{i})^{5} is equal to & (q) & 4 (C) & 4 \prod i = 1 (z_{i} + 2) is equal to & (r) & 1 (D) & Least value of [| z_{1} + z_{2} |] is & (s) & 11 \end{matrix}

where []

represents greatest integer function

Which of the following is the correct combination

Q. Solutions of the equation

x | x + 1 | + 1 = 0

are

If $k > 1$ and the determinant of the matrix $A^{2}$ is $k^{2}$ , then $| α | =$

$A = [\begin{matrix} k & k α & α \\ 0 & α & k α \\ 0 & 0 & k \end{matrix}]$

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The solutions of the equation1+sin2xsin2xsin2xcos2x1+cos2xcos2x4sin2x4sin2x1+4sin2x=0,0<x<π are:

The correct option is B 7π12,11π12Explanation for the correct optionR1→R1+R2221cos2x1+cos2xcos2x4sin2x4sin2x1+4sin2x=0C1→C1-C2021-11+cos2xcos2x04sin2x1+4sin2x=0∴2+8sin2x-4sin2x=0⇒sin2x=-12⇒x=7π12,11π12Hence the correct option is option(b) i.e. 7π12,11π12

The solutions of the equation $|\begin{matrix} 1 + \sin^{2} x & \sin^{2} x & \sin^{2} x \\ \cos^{2} x & 1 + \cos^{2} x & \cos^{2} x \\ 4 \sin 2 x & 4 \sin 2 x & 1 + 4 \sin 2 x \end{matrix}| = 0, (0 < x < π)$ are: