(C) Given equations are $\cos \theta = \frac{1}{\sqrt{2}}$ and $\tan \theta = -1$ for $\theta \in [0, 2\pi]$.For $\cos \theta = \frac{1}{\sqrt{2}}$,th

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(C) Given equations are $\cos \theta = \frac{1}{\sqrt{2}}$ and $\tan \theta = -1$ for $\theta \in [0, 2\pi]$.For $\cos \theta = \frac{1}{\sqrt{2}}$,the possible values of $\theta$ are $\frac{\pi}{4}$ and $\frac{7\pi}{4}$.For $\tan \theta = -1$,the possible values of $\theta$ are $\frac{3\pi}{4}$ and $\frac{7\pi}{4}$.The common value satisfying both equations is $\theta = \frac{7\pi}{4}$.

Class 11 Mathematics Trigonometrical Equations Mix Examples-Trigonometrical Equations and Inequations, Properties of Triangles, Height and Distance

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The value of $\theta$,satisfying both the equations $\cos \theta = \frac{1}{\sqrt{2}}$ and $\tan \theta = -1$ in the interval $[0, 2\pi]$,is

MHT CET 2022 All MHT CET

A
$\frac{\pi}{4}$
B
$\frac{5\pi}{4}$
C
$\frac{7\pi}{4}$
D
$\frac{3\pi}{4}$

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Trigonometrical Equations

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Mix Examples-Trigonometrical Equations and Inequations, Properties of Triangles, Height and Distance

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$ABCD$ is a trapezium such that $AB$ and $CD$ are parallel and $BC \perp CD$. If $\angle ADB = \theta$,$BC = p$ and $CD = q$,then $AB$ is equal to:

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In a $\Delta ABC$,if $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$ and the side $a = 2$,then the area of the triangle is:

MediumIIT 1993

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In a triangle $ABC$ with usual notations,if $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$,then the triangle is equilateral. If the side length is $a = \sqrt{6}$,find the area of the triangle.

MediumMHT CET 2020

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If $x = \frac{n\pi}{2}$ satisfies the equation $\sin \frac{x}{2} - \cos \frac{x}{2} = 1 - \sin x$ and the inequality $\left| \frac{x}{2} - \frac{\pi}{2} \right| \le \frac{3\pi}{4}$,then:

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

DifficultIIT 2014

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The value of $\theta$,satisfying both the equations $\cos \theta = \frac{1}{\sqrt{2}}$ and $\tan \theta = -1$ in the interval $[0, 2\pi]$,is

Similar Questions

$ABCD$ is a trapezium such that $AB$ and $CD$ are parallel and $BC \perp CD$. If $\angle ADB = \theta$,$BC = p$ and $CD = q$,then $AB$ is equal to:

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