How do you solve $ \cos x + 1 = \sin x $ ?

Hint: $ \sin x $ is a trigonometric function. In a right-angle triangle ABC, it is defined as the ratio between perpendicular and hypotenuse to angle $ a $ . $ \cos x $ is also a trigonometric function. In a right-angle triangle ABC, it is defined as the ratio between base and hypotenuse to angle $ b $ . We know that the value of $ x $ remains between $ {0^ \circ } $ and $ {360^ \circ } $ . The domain of both $ \sin x $ and $ \cos x $ is $ \left( { - \infty ,\infty } \right) $ and range is between $ \left[ { - 1,1} \right] $ .

Complete step-by-step answer:
Given trigonometric function is $ \cos x + 1 = \sin x $ .
First, let us bring $ \sin x $ on the left-hand side which can be done by subtracting $ \sin x $ from both the right and left-hand side.
 $
  \cos x + 1 - \sin x = \sin x - \sin x \\
  \cos x - \sin x + 1 = 0 \;
  $
Let’s take $ 1 $ on the right-hand side which can be done in the same way by which we transferred $ \sin x $ from right-hand side to left-hand side i.e., by subtracting $ 1 $ from both sides.
 $
  \cos x - \sin x + 1 - 1 = 0 - 1 \\
  \cos x - \sin x = - 1 \;
  $
Now, let us take minus (-) symbol common from all the terms.
 $ - \left( {\sin x - \cos x} \right) = - 1 $
Minus symbol cancels off from both sides and we get,
 $ \sin x - \cos x = 1 $ ----(1)
Now, we have to multiply (1) with $ \dfrac{{\sqrt 2 }}{2} $ and we get,
 $
   \Rightarrow \sin x \times \dfrac{{\sqrt 2 }}{2} - \cos x \times \dfrac{{\sqrt 2 }}{2} = 1 \times \dfrac{{\sqrt 2 }}{2} \\
   \Rightarrow \sin x \times \dfrac{{\sqrt 2 }}{2} - \cos x \times \dfrac{{\sqrt 2 }}{2} = \dfrac{{\sqrt 2 }}{2} \\
   \Rightarrow \sin \left( {x - \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{2} \\
   \Rightarrow x - \dfrac{\pi }{4} = \dfrac{\pi }{4} + 2k\pi ,\;for\;any\;\mathbb{Z}\;k \\
   \Rightarrow x = \dfrac{\pi }{2} + 2k\pi \;
  $
Now, let us keep $ \sin x = 0 $
 $ x = \arcsin \left( 0 \right) $
We know that the value of $ \arcsin \left( 0 \right) $ is $ 0 $ .
 $ x = 0 $
Trigonometric function $ \sin x $ is positive in the first and second quadrants. In order to find a second solution, we are supposed to subtract the reference angle from $ \pi $ .
 $ x = \pi - 0 $
The period of the function is calculated by using $ \dfrac{{2\pi }}{{|b|}} $ . By replacing $ b $ with $ 1 $ we get,
 $ \dfrac{{2\pi }}{{|1|}} = 2\pi $
So, we now know that the period of $ \sin x $ is $ 2\pi $ which means that values will repeat after every $ 2\pi $ radians in both the directions.
 $ x = 2k\pi ,\pi + 2k\pi ,\;for\;any\;\mathbb{Z}\;k $

Note: $ \cos x $ is zero at multiples of $ 90 $ degrees and $ \sin x $ is zero at $ 0 $ degrees and multiples of $ 180 $ degrees, so they are never zero for the same $ x $ . Moreover, $ \sin x $ and $ \cos x $ are equal at $ \dfrac{\pi }{4} $ only. The sine function graphs are known as sine waves. The cosine function graph is just like sine waves but it starts from $ 1 $ and falls till $ - 1 $ .