Solution.
sin ( x ) − cos ( x ) = − 0.3 , \sin(x) - \cos(x) = -0.3, sin ( x ) − cos ( x ) = − 0.3 , sin ( x ) − cos ( x ) = − 2 ( cos ( x ) 2 − sin ( x ) 2 ) = − 2 ( cos ( π 4 ) cos ( x ) − sin ( π 4 ) sin ( x ) ) = − 2 cos ( π 4 + x ) , \sin(x) - \cos(x) = -\sqrt{2} \left( \frac{\cos(x)}{\sqrt{2}} - \frac{\sin(x)}{\sqrt{2}} \right) = -\sqrt{2} \left( \cos \left( \frac{\pi}{4} \right) \cos(x) - \sin \left( \frac{\pi}{4} \right) \sin(x) \right) = -\sqrt{2} \cos \left( \frac{\pi}{4} + x \right), sin ( x ) − cos ( x ) = − 2 ( 2 cos ( x ) − 2 sin ( x ) ) = − 2 ( cos ( 4 π ) cos ( x ) − sin ( 4 π ) sin ( x ) ) = − 2 cos ( 4 π + x ) , So − 2 cos ( π 4 + x ) = − 3 10 . \text{So } -\sqrt{2} \cos \left( \frac{\pi}{4} + x \right) = -\frac{3}{10}. So − 2 cos ( 4 π + x ) = − 10 3 .
Divide both sides by − 2 -\sqrt{2} − 2
cos ( π 4 + x ) = 3 10 2 , \cos \left( \frac{\pi}{4} + x \right) = \frac{3}{10\sqrt{2}}, cos ( 4 π + x ) = 10 2 3 ,
For cos ( x ) = a , x = ± arccos ( a ) + 2 π n , n ∈ Z \cos(x) = a, x = \pm \arccos(a) + 2\pi n, n \in \mathbb{Z} cos ( x ) = a , x = ± arccos ( a ) + 2 πn , n ∈ Z
π 4 + x = ± arccos ( 3 10 2 ) + 2 π n , n ∈ Z . \frac{\pi}{4} + x = \pm \arccos \left( \frac{3}{10\sqrt{2}} \right) + 2\pi n, n \in \mathbb{Z}. 4 π + x = ± arccos ( 10 2 3 ) + 2 πn , n ∈ Z .
Subtract π 4 \frac{\pi}{4} 4 π
x = ± arccos ( 3 10 2 ) − π 4 + 2 π n , n ∈ Z . x = \pm \arccos \left( \frac{3}{10\sqrt{2}} \right) - \frac{\pi}{4} + 2\pi n, n \in \mathbb{Z}. x = ± arccos ( 10 2 3 ) − 4 π + 2 πn , n ∈ Z .
Answer.
x = ± arccos ( 3 10 2 ) − π 4 + 2 π n , n ∈ Z . x = \pm \arccos \left( \frac{3}{10\sqrt{2}} \right) - \frac{\pi}{4} + 2\pi n, n \in \mathbb{Z}. x = ± arccos ( 10 2 3 ) − 4 π + 2 πn , n ∈ Z . 
#339914 on Dec 2023