If ${\log _2}\sin x - {\log _2}\cos x - {\log _2}\left( {1 - {{\tan }^2}x} \right) = - 1,$ then $x = $

$\frac{{n{\rm{\pi }}}}{2} + \frac{{\rm{\pi }}}{8},\;n \in Z$

$n{\rm{\pi }} - \frac{{\rm{\pi }}}{8},n \in Z$

$\frac{{n{\rm{\pi }}}}{4} + \frac{{\rm{\pi }}}{2},n \in Z$

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Analyzing +4 | -1

{\log _2}\sin x - {\log _2}\cos x - {\log _2}\left( {1 - {{\tan }^2}x} \right) = - 1,

then

x =

View answer and explanation

Correct option: A

\frac{{n\mathrm{\pi }}}{2} + \frac{\mathrm{\pi }}{8},\;n \in Z

Option A (

\frac{{n\mathrm{\pi }}}{2} + \frac{\mathrm{\pi }}{8},\\;n \in Z

) is correct because solving the trigonometric logarithmic equation leads to this general solution for x.

Practice All Chapter MCQs →

A=\{1,2,3\}, B=\{3,4\}, C=\{4,5,6\}

{A} \cup({B} \cap {C})

A-(B \cap C)

(B \cup C)-A

a,b,c

a,x,b

b,y,c

{x^2},{b^2},{y^2}

P(x)=1-x+x^2-x^3+\ldots+x^{2012}

y

Q(y)=a_0+a_1 y+\ldots+a_{2012} y^{2012}

y=x-2

\sum\limits_{{i=0}}^{2012} a_i

\log \left(\frac{5 c}{a}\right), \log \left(\frac{3 b}{5 c}\right)

\log \left(\frac{a}{3 b}\right)

A P

a, b

G P

a, b, c

\frac{3}{4}