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Question: How do you solve for \(x\) in \(3\sin 2x=\cos 2x\) for the interval \(0\le x<2\pi \)?...

How do you solve for xx in 3sin2x=cos2x3\sin 2x=\cos 2x for the interval 0x<2π0\le x<2\pi ?

Explanation

Solution

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 sinθ\sin \theta is sinθ=ah\sin \theta =\dfrac{a}{h} where θ\theta is the angle theta, aais the length of opposite side andhh is the length of the hypotenuse. The formula of finding the cosθ\cos \theta iscosθ=bh\cos \theta =\dfrac{b}{h} where bb is the base of the triangle. In the above question we have been given 3sin2x=cos2x3\sin 2x=\cos 2x.
To solve this we will use the formula of trigonometric which is as:
tanx=sinxcosx\Rightarrow \tan x=\dfrac{\sin x}{\cos x}
Now we can this 3sin2x=cos2x3\sin 2x=\cos 2xas:
3.sin2xcos2x=1\Rightarrow 3.\dfrac{\sin 2x}{\cos 2x}=1
Now dividing the both sides of the above equation by 33 we get,
sin2xcos2x=13\Rightarrow \dfrac{\sin 2x}{\cos 2x}=\dfrac{1}{3}
Now by using the tanx=sinxcosx\tan x=\dfrac{\sin x}{\cos x}, we get
tan2x=13\Rightarrow \tan 2x=\dfrac{1}{3}
Now we will simply multiply the both sides of equation by tan1{{\tan }^{-1}} then, we gettan1tan2x=tan1(13)\Rightarrow {{\tan }^{-1}}\tan 2x={{\tan }^{-1}}\left( \dfrac{1}{3} \right)
Now the tan1tan{{\tan }^{-1}}\tan both get cancel out and we know the value of tan1(13){{\tan }^{-1}}\left( \dfrac{1}{3} \right) is 0.3210.321 by putting all these in above equation we get
2x=0.321\Rightarrow 2x=0.321
Now divide the both sides of equation by 22, we get
x=0.3212\Rightarrow x=\dfrac{0.321}{2}
By using calculator we got the value of xx is,
x=0.160\Rightarrow x=0.160
Hence the value of xxis 0.1600.160 for the interval 0x<2π0\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.