Question

Solve $\sin 3x=\cos 3x$ for ${{0}^{\circ }} < x < {{360}^{\circ }}$ ?

Answer 1

Hint: Here we have been given an equation with trigonometric function in it and we have to find the value of $x$ whose range is given. Firstly we will take the right hand side value to the left and side and using trigonometric relation convert the fraction into non-fraction value. Then we will take the trigonometric function on the right side and solve the value obtained to get our desired answer.

Complete step by step answer:
We have to solve the below equation:
$\sin 3x=\cos 3x$….$\left( 1 \right)$
For ${{0}^{\circ }} < x < {{360}^{\circ }}$
Now take the right hand side value to the left hand side in equation (1),
$\Rightarrow \dfrac{\sin 3x}{\cos 3x}=1$
Now as we know the relation that $\dfrac{\sin A}{\cos A}=\tan A$ using it above where $A=3x$ we get,
$\Rightarrow \tan 3x=1$
Take the trigonometric function on the right side as follows:
$\Rightarrow 3x={{\tan }^{-1}}1$
We know $\tan \left( \dfrac{\pi }{4}+k\pi \right)=1$ for all $k\in Z$ substituting it above,
$\Rightarrow 3x={{\tan }^{-1}}\tan \left( \dfrac{\pi }{4}+k\pi \right)$
As ${{\tan }^{-1}}\tan x=x$ we get,
$\Rightarrow 3x=\left( \dfrac{\pi }{4}+k\pi \right)$
$\Rightarrow x=\dfrac{1}{3}\left( \dfrac{\pi }{4}+k\pi \right)$
 So we get,
$\Rightarrow x=\dfrac{\pi }{12}+\dfrac{k\pi }{3}$
As $\pi ={{180}^{\circ }}$ we get,
$\Rightarrow x=\dfrac{{{180}^{\circ }}}{12}+\dfrac{{{180}^{\circ }}}{3}k$
$\Rightarrow x={{15}^{\circ }}+{{60}^{\circ }}k$ $\forall \,k\in Z$
As we know ${{0}^{\circ }} < x < {{360}^{\circ }}$
So we will let $k$ values till our $x$ lies in the above range.
Let $k=1$
$\Rightarrow x={{15}^{\circ }}+{{60}^{\circ }}\times 1$
$\Rightarrow x={{75}^{\circ }}$
Let $k=2$
$\Rightarrow x={{15}^{\circ }}+{{60}^{\circ }}\times 2$
$\Rightarrow x={{135}^{\circ }}$
Let $k=3$
$\Rightarrow x={{15}^{\circ }}+{{60}^{\circ }}\times 3$
$\Rightarrow x={{195}^{\circ }}$
Let $k=4$
$\Rightarrow x={{15}^{\circ }}+{{60}^{\circ }}\times 4$
$\Rightarrow x={{255}^{\circ }}$
Let $k=5$
$\Rightarrow x={{15}^{\circ }}+{{60}^{\circ }}\times 5$
$\Rightarrow x={{315}^{\circ }}$
Now if $k=6$ our $x>{{360}^{\circ }}$
So we got the value of $x={{75}^{\circ }},{{135}^{\circ }},{{195}^{\circ }},{{255}^{\circ }},{{315}^{\circ }}$
Hence the solution of $\sin 3x=\cos 3x$ for ${{0}^{\circ }} < x < {{360}^{\circ }}$ is $x={{75}^{\circ }},{{135}^{\circ }},{{195}^{\circ }},{{255}^{\circ }},{{315}^{\circ }}$.

Note:
As the range of the unknown variable is given we have further simplified our answer by taking different values of $k$ . If the range was not given our answer would be the general solution which is true for all values of $k$ . The relation between the six trigonometric functions is very useful in such questions as it removes the long method of calculation. Also when we take the trigonometric function on the other side it becomes an inverse function and to solve that we need to use the inverse function properties and formulas.