Question

If $\sin x + \cos x - 2 \sqrt { 2 } \sin x \cos x = 0$ then the general solution of x is

• A. $x = 2 n \pi + \frac { \pi } { 4 }$
• B. $x = n \pi + ( - 1 ) ^ { n } \frac { \pi } { 6 } - \frac { \pi } { 4 }$
• C. Both (1) and (2)
• D. None of these

Answer 1

Answer: (C)

Both (1) and (2)

Answer 2

Let $( \sin x + \cos x ) = t$ and using the equation

$\sin x \cdot \cos x = \frac { t ^ { 2 } - 1 } { 2 }$ we get $t - 2 \sqrt { 2 } \left( \frac { t ^ { 2 } - 1 } { 2 } \right) = 0$

$\sqrt { 2 } t ^ { 2 } - t - \sqrt { 2 } = 0$

The numbers $t _ { 1 } = \sqrt { 2 } , t _ { 2 } = - \frac { 1 } { \sqrt { 2 } }$ are roots of this quadratic

equation.

Thus the solution of the given equation reduces to the solution of two trigonometrical equation;

$\sin x + \cos x = - \frac { 1 } { \sqrt { 2 } }$

or$\frac { 1 } { \sqrt { 2 } } \sin x + \frac { 1 } { \sqrt { 2 } } \cos x = 1$ or $\frac { 1 } { \sqrt { 2 } } \sin x + \frac { 1 } { \sqrt { 2 } } \cos x = - \frac { 1 } { 2 }$

or $\sin x \cdot \cos \frac { \pi } { 4 } + \sin \frac { \pi } { 4 } \cos x = 1$or

$\sin x \cos \frac { \pi } { 4 } + \sin \frac { \pi } { 4 } \cos x = - \frac { 1 } { 2 }$

$\sin \left( x + \frac { \pi } { 4 } \right) = 1$ or $\sin \left( x + \frac { \pi } { 4 } \right) = - \frac { 1 } { 2 }$

$x + \frac { \pi } { 4 } = ( 4 n + 1 ) \frac { \pi } { 2 }$ or $x + \frac { \pi } { 4 } = n \pi + ( - 1 ) ^ { n } \cdot \left( \frac { \pi } { 6 } \right)$

$x = 2 n \pi + \frac { \pi } { 4 }$ or $x = n \pi + ( - 1 ) ^ { n } \frac { \pi } { 6 } - \left( \frac { \pi } { 4 } \right)$ .