Question

How do you solve $\sin x = 1 - \cos x$ over the interval $(0,2pi]$ ?

Answer 1

Here we are given an equation that is $\sin x = 1 - \cos x$ and we are asked to solve it over the interval $(0,2pi]$. After seeing the equation we can judge that it is a trigonometric equation that is the equation containing the trigonometric quantities and for solving this we would the trigonometric identity that is ${\sin ^2}x + {\cos ^2}x = 1$ and would further simplify it and solve it then equate it for getting the values of the $x$ and we would find the generalized solutions.

Complete step by step solution:
Here we are given a trigonometric equations that is $\sin x = 1 - \cos x$ and we are asked to solve it over the interval given that is $(o,2pi]$ so solving the equation can be done as –
Squaring the both side of the equation that comes out to be as –

$${(\sin x)^2} = {(1 - \cos x)^2} \\\ {\sin ^2}x = 1 + {\cos ^2}x - 2\cos x \\\$$

Now using the trigonometric identity that is ${\sin ^2}x + {\cos ^2}x = 1$ in above equation and simplifying it further-

$${\sin ^2}x = 1 + {\cos ^2}x - 2\cos x \\\ 0 = 1 + {\cos ^2}x - {\sin ^2}x - 2\cos x \\\ 0 = 1 + {\cos ^2}x - \left( {1 - {{\cos }^2}x} \right) - 2\cos x \\\$$

Using$\sin {x^2} = \left( {1 - {{\cos }^2}x} \right)$ from the trigonometric identity ${\sin ^2}x + {\cos ^2}x = 1$ in above equation
Now further simplifying the equation we get –
$0 = 2{\cos ^2}x - 2\cos x$
Now taking the $2$ common from the RHS and solving it further we get-

$$0 = {\cos ^2}x - \cos x \\\ 0 = \cos x\left( {\cos x - 1} \right) \\\$$

Now equating the both the terms with the zero we get –

$$\left( {\cos x - 1} \right) = 0 \\\ \cos x = 1 \\\$$

Now the values of the x can be $\dfrac{\pi }{2},\dfrac{{3\pi }}{2}$ in the interval stated in the question that is $(0,2pi]$
Also $\cos x = 0$ so x can be $0$ in the interval stated in the question that is $(0,2pi]$
Now we would put the values of the x that is in $\dfrac{\pi }{2},\dfrac{{3\pi }}{2}$ and $0$ in the above given trigonometric equation $\sin x = 1 - \cos x$ to check which one of the solution satisfies the equation
Now putting x as $\dfrac{\pi }{2}$ in the equation $\sin x = 1 - \cos x$ we get –

\sin \dfrac{\pi }{2} = 1 - \cos \dfrac{\pi }{2} \\\ 1 = 1 - 0 \\\ 1 = 1 \\\ $$as the values of the $$\sin \dfrac{\pi }{2},\cos \dfrac{\pi }{2}$$ are $$1,0$$ respectively So the solution $$\dfrac{\pi }{2}$$ satisfies the equation so it is answer to the equation Now checking for the $$\dfrac{{3\pi }}{2}$$ we get –

\sin \dfrac{{3\pi }}{2} = 1 - \cos \dfrac{{3\pi }}{2} \\
- 1 \ne 1 - 0 \\
- 1 \ne 1 \\
$as the values of the$\sin \dfrac{{3\pi }}{2},\cos \dfrac{{3\pi }}{2}$are$ - 1,0$respectively So the solution$\dfrac{{3\pi }}{2}$does not satisfy the equation so it is not the answer of the equation Now checking for the$0$$ we get –

\sin 0 = 1 - \cos 0 \\\ 0 = 1 - 1 \\\ 0 = 0 \\\ $$ as the values of the $$\sin 0,\cos 0$$ are $$0,1$$ respectively So the solution $$0$$ satisfies the equation so it is answer to the equation **The values satisfying the equation $$\sin x = 1 - \cos x$$ are $$\dfrac{\pi }{2}$$ and $$0$$ in the interval stated in the question that is $$(o,2pi]$$** **Note:** While solving such kinds of the questions one should have the knowledge about the different trigonometric identities and the right way to use them and the values of the trigonometric components at the various angles that helps in the solving of the question easily.