Question

How do you solve $\cos x - 4 = \sin x - 4$ for $0 \leqslant x \leqslant 2\pi $?

Answer 1

Hint: First we will evaluate the right-hand of equation and then further the left-hand side of the equation. We will use the relation $\tan \left( {\dfrac{\pi }{4}} \right) = 1$. Then we will try to factorise and simplify the terms so that the left-hand side matches the right-hand side. Then finally evaluate the solution which lies within the given interval.

Complete step by step answer:
We will start solving this question by adding the term $4$ to both the sides of the equation.
$
  \,\,\,\,\,\,\,\cos x - 4 = \sin x - 4 \\
  \cos x - 4 + 4 = \sin x - 4 + 4 \\
 $
Now, if we simplify the equation further we will get,
$\cos x = \sin x$
Now if we know that it is either $\cos x = \sin x$ or $\tan x = 1$
We also know that, $\tan \left( {\dfrac{\pi }{4}} \right) = 1$ and $\tan \left( {\pi + \dfrac{\pi }{4}} \right) = 1$ or we can say $\tan \left( {\dfrac{{5\pi }}{4}} \right) = 1$
Hence, the values of $x$ will be $\dfrac{\pi }{4},\dfrac{{5\pi }}{4}$.
Now, we know that the given interval is $0 \leqslant x \leqslant 2\pi $ for all values of $x$.
Since, the terms $\dfrac{\pi }{4},\dfrac{{5\pi }}{4}$ lie within the range $[0,2\pi ]$ therefore, both the solutions are feasible.

Hence, the solution of the expression $\cos x - 4 = \sin x - 4$ within the range is $\dfrac{\pi }{4},\dfrac{{5\pi }}{4}$.

Note: While choosing the side to solve, always choose the side where you can directly apply the trigonometric identities. Also, remember the trigonometric identities ${\sin ^2}x + {\cos ^2}x = 1$ and $\cos 2x = 2{\cos ^2}x - 1$. While applying the double angle identities, first choose the identity according to the terms you have then choose the terms from the expression involving which you are using the double angle identities. While modifying any identity make sure that when you back trace the identity, you get the same original identity. Also, remember that the range of $\cos $ function is from $ - 1\,$ to $ + 1$ and the range of $\sin $ function is also from $ - 1\,$ to $ + 1$.