Question

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

Answer 1

In this question, we are given a trigonometric equation and we need to find the general value of x which will satisfy the equation. For this we will first factorize it and then find an individual trigonometric function equal to some constant. Using trigonometric ratio tables, we will find a particular value of x and then use it to find the general value of x. We will use following formulas,
(I) tanx is equal to $\dfrac{\sin x}{\cos x}$.
(II) If $\tan x=\tan \alpha$ then $x=n\pi \pm \alpha ,n\in z$.
(III) If $\cos x=\cos \alpha$ then $x=2n\pi \pm \alpha ,n\in z$.

Complete step by step solution:
Here we are given the trigonometric equation as $\sin x-\cos x-\tan x=-1$.
We need to find the value of x which satisfies this equation. For this let us simplify this equation.
Let us bring 1 from the right side of the equation to the left side we get $\sin x-\cos x-\tan x+1=0$.
Now let us change tanx into form of sinx and cosx using the formula $\tan x=\dfrac{\sin x}{\cos x}$ we get $\sin x-\cos x-\dfrac{\sin x}{\cos x}+1=0$.
Now let us take the negative sign common from the last two terms: we get $\sin x-\cos x-\left( \dfrac{\sin x}{\cos x}-1 \right)=0$.
Now let us take LCM of cosx inside the bracket term, we have the equation as $\left( \sin x-\cos x \right)-\left( \dfrac{\sin x-\cos x}{\cos x} \right)=0$.
Now to remove cosx from the denominator let us multiply both sides of the equation by cosx we get $\cos x\left( \left( \sin x-\cos x \right)-\left( \dfrac{\sin x-\cos x}{\cos x} \right) \right)=0\times \cos x$.
Right side of the equation will be zero again. Separating cosx for both terms in the left side of the equation we get $\left( \sin x-\cos x \right)\cos x-\left( \dfrac{\sin x-\cos x}{\cos x} \right)\cos x=0$.
Cancelling cosx from the numerator and the denominator in the second term we get $\left( \sin x-\cos x \right)\cos x-\left( \sin x-\cos x \right)=0$.
Now we can factorize this equation by taking $\left( \sin x-\cos x \right)$ from both terms on the left side we get $\left( \sin x-\cos x \right)\left( \cos x-1 \right)=0$.
Since both factors are equal to 0. So any one of them can be zero such that the equation satisfies. Let us take both factors as zero one by one.
(1) sinx - cosx = 0.
Taking cosx to the other side we get sinx = cosx.
As we know cosx cannot be equal to zero because then tanx will be undefined in original equation so, dividing both sides by cosx we get $\dfrac{\sin x}{\cos x}=\dfrac{\cos x}{\cos x}$.
Using $\tan x=\dfrac{\sin x}{\cos x}$ on the left side and $\dfrac{\cos x}{\cos x}=1$ on the right side we get tanx = 1.
From the trigonometric ratio table we know that $\tan \dfrac{\pi }{4}=1$. So we can write the above equation as $\tan x=\tan \dfrac{\pi }{4}$.
Now we know that if $\tan x=\tan \alpha$ then $x=n\pi \pm \alpha ,n\in z$.
So here we have $\tan x=\tan \dfrac{\pi }{4}\Rightarrow x=n\pi \pm \dfrac{\pi }{4},n\in z$.
So some values of x are $n\pi \pm \dfrac{\pi }{4},n\in z$.
(2) cosx - 1 = 0.
Taking 1 to the other side we get cosx = 1.
From the trigonometric ratio table we know that cos0 = 1. Therefore we get the equation as cosx = cos0.
Now we know that if $\cos x=\cos \alpha$ then $x=2n\pi \pm \alpha ,n\in z$.
So here we have cosx = cos0.
$x=2n\pi \pm 0,n\in z\Rightarrow x=2n\pi ,n\in z$.
Therefore some values of x are $x=2n\pi ,n\in z$.
From (1) and (2) we see that the values of x are $n\pi \pm \dfrac{\pi }{4},n\in z$ and $x=2n\pi ,n\in z$.
This is our required solution of n.

Note: Students should keep in mind the properties of trigonometric function for solving this question. Note that we can put the value of n as any integer and we will get the value of x for which the equation satisfies.