General solution of $\sin \,\,x + \cos \,\,x = \begin{array}{*{20}{c}} {min}\\ {a \in R} \end{array}\,\,\left\{ {1,\,\;{a^2} - 4a + 6} \right\}$ is

$\frac{{n\pi }}{2} + {\left( { - 1} \right)^n}\frac{\pi }{4}$

$2\,n\pi + {\left( { - 1} \right)^n}\,\,\frac{\pi }{4}$

$n\pi + {\left( { - 1} \right)^{n + 1}}\,\,\frac{\pi }{4}$

JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

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Application +4 | -1

General solution of

\sin \,\,x + \cos \,\,x = \begin{array}{cccccccccccccccccccc} {min}\\ {a \in R} \end{array}\,\,\left\{ {1,\,\;{a^2} - 4a + 6} \right\}

View answer and explanation

Correct option: D

n\pi + {\left( { - 1} \right)^n}\,\,\frac{\pi }{4} - \frac{\pi }{4}

Option D (

n\pi + {(- 1)^n}\,\,\frac{\pi }{4} - \frac{\pi }{4}

) is correct because it represents the general solution of the given trigonometric equation.

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