Question

How do you solve for $x$ in $3\sin 2x=\cos 2x$ for the interval $0\le x<2\pi $?

Answer 1

Hint: Before solving the above question let's discuss trigonometric functions. In mathematics, trigonometric functions are the real functions which relate an angle of a right-angled triangle to ratios of two side lengths. They are widely used in sciences that are related to geometry, such as navigation, solid mechanics, and many others

Complete step by step solution:
In the above question we have been given sin function and cos function. The formula of finding the $\sin \theta $ is $\sin \theta =\dfrac{a}{h}$ where $\theta $ is the angle theta, $a$is the length of opposite side and$h$ is the length of the hypotenuse. The formula of finding the $\cos \theta $ is$\cos \theta =\dfrac{b}{h}$ where $b$ is the base of the triangle. In the above question we have been given $3\sin 2x=\cos 2x$.
To solve this we will use the formula of trigonometric which is as:
$\Rightarrow \tan x=\dfrac{\sin x}{\cos x}$
Now we can this $3\sin 2x=\cos 2x$as:
$\Rightarrow 3.\dfrac{\sin 2x}{\cos 2x}=1$
Now dividing the both sides of the above equation by $3$ we get,
$\Rightarrow \dfrac{\sin 2x}{\cos 2x}=\dfrac{1}{3}$
Now by using the $\tan x=\dfrac{\sin x}{\cos x}$, we get
$\Rightarrow \tan 2x=\dfrac{1}{3}$
Now we will simply multiply the both sides of equation by ${{\tan }^{-1}}$ then, we get$\Rightarrow {{\tan }^{-1}}\tan 2x={{\tan }^{-1}}\left( \dfrac{1}{3} \right)$
Now the ${{\tan }^{-1}}\tan $ both get cancel out and we know the value of ${{\tan }^{-1}}\left( \dfrac{1}{3} \right)$ is $0.321$ by putting all these in above equation we get
$\Rightarrow 2x=0.321$
Now divide the both sides of equation by $2$, we get
$\Rightarrow x=\dfrac{0.321}{2}$
By using calculator we got the value of $x$ is,
$\Rightarrow x=0.160$
Hence the value of $x$is $0.160$ for the interval $0\le x<2\pi $.

Note: We can go wrong by using the wrong trigonometric formula. Trigonometric has many formulas and concepts. so always keep those formulas in mind before solving the trigonometric question.