Question

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

Answer 1

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 $- 1$ .