Let $P=\{\theta:\sin\theta-\cos\theta=\sqrt{2}\cos\theta\}$ and $Q=\{\theta:\sin\theta+\cos\theta=\sqrt{2}\sin\theta\}$ be two sets. Then 

$P=\left\{ \theta :\sin { \theta -\cos { \theta } =\sqrt { 2 } \cos { \theta } } \right\}$

$\sin { \theta -\cos { \theta } =\sqrt { 2 } \cos { \theta } }$

Dividing by $\cos { \theta }$, we get

$\tan { \theta } -1=\sqrt { 2 }$

$\tan { \theta } =\sqrt { 2 } + 1$

$\theta =\arctan { (\sqrt { 2 } +1) }$

$Q=\left\{ \theta :\sin { \theta +\cos { \theta } =\sqrt { 2 } \sin { \theta } } \right\}$

$\sin { \theta } +\cos { \theta } =\sqrt { 2 } \sin { \theta }$

$\cos { \theta } =\sqrt { 2 } \sin { \theta } -\sin { \theta }$

$\cos { \theta } =(\sqrt { 2 } -1)\sin { \theta }$

$\tan { \theta } =\dfrac { 1 }{ \sqrt { 2 } -1 }$

On rationalising, we get

$\tan { \theta } =\sqrt { 2 } +1$

$\theta =\arctan { (\sqrt { 2 } +1) }$

Hence, sets P and Q both have same values. So $P = Q$.