Question
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 x in 3sin2x=cos2x for the interval 0≤x<2π?
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 is sinθ=ha where θ is the angle theta, ais the length of opposite side andh is the length of the hypotenuse. The formula of finding the cosθ iscosθ=hb where b is the base of the triangle. In the above question we have been given 3sin2x=cos2x.
To solve this we will use the formula of trigonometric which is as:
⇒tanx=cosxsinx
Now we can this 3sin2x=cos2xas:
⇒3.cos2xsin2x=1
Now dividing the both sides of the above equation by 3 we get,
⇒cos2xsin2x=31
Now by using the tanx=cosxsinx, we get
⇒tan2x=31
Now we will simply multiply the both sides of equation by tan−1 then, we get⇒tan−1tan2x=tan−1(31)
Now the tan−1tan both get cancel out and we know the value of tan−1(31) is 0.321 by putting all these in above equation we get
⇒2x=0.321
Now divide the both sides of equation by 2, we get
⇒x=20.321
By using calculator we got the value of x is,
⇒x=0.160
Hence the value of xis 0.160 for the interval 0≤x<2π.
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.