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Question: How do you find all solutions of the equations \(\cos x+\sin x\tan x=2\) in the interval \([0,2\pi )...

How do you find all solutions of the equations cos⁡x+sin⁡xtan⁡x=2\cos x+\sin x\tan x=2 in the interval [0,2π)[0,2\pi ) ?
(a) By checking random values
(b) Trying drawing the graph
(c) Simplifying the equation
(d) None of these

Explanation

Solution

We are given the trigonometric equation cos⁡x+sin⁡xtan⁡x=2\cos x+\sin x\tan x=2. To get the solution as x, we need to simplify the equation in terms of sinx or cosx. Thus we are getting the value of sinx or cos x as a given number. Thus, we can find the value of x in the general form and that is what we are looking for.

Complete step-by-step answer:
According to the question, we have our equation given as, cos⁡x+sin⁡xtan⁡x=2\cos x+\sin x\tan x=2
Now, putting, tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x} , we are getting,
⇒cos⁡x+sin⁡x.sin⁡xcos⁡x=2\Rightarrow \cos x+\sin x.\dfrac{\sin x}{\cos x}=2
Multiplying both sinx terms in the numerator,
⇒cos⁡x+sin⁡2xcos⁡x=2\Rightarrow \cos x+\dfrac{{{\sin }^{2}}x}{\cos x}=2
Now again, we know as per the trigonometric identities,sin⁡2x=1−cos⁡2x{{\sin }^{2}}x=1-{{\cos }^{2}}x , because, sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1.
So, now if we put the value of sin⁡2x{{\sin }^{2}}xin our equation, we will get a equation with numbers and cosx.
⇒cos⁡x+1−cos⁡2xcos⁡x=2\Rightarrow \cos x+\dfrac{1-{{\cos }^{2}}x}{\cos x}=2
Multiplying both sides with cos⁡x\cos x , we will get rid of the denominator,
⇒cos⁡2x+1−cos⁡2x=2cos⁡x\Rightarrow {{\cos }^{2}}x+1-{{\cos }^{2}}x=2\cos x
After more simplification and then cancelling out,
⇒2cos⁡x=1\Rightarrow 2\cos x=1
So, now, we get the value of cos⁡x\cos x be, cos⁡x=12\cos x=\dfrac{1}{2} .
Hence, according to the trigonometric table, the value of cosine function in the point π3\dfrac{\pi }{3} is 12\dfrac{1}{2} . If we try to use the general solution of cosine function, we will get the solution of x as,
x=2nπ±π3,n∈Zx=2n\pi \pm \dfrac{\pi }{3},n\in \mathbb{Z}
As, the general solution of cos⁡x=cos⁡y\cos x=\cos y, gives us, x=2nπ±y,n∈Zx=2n\pi \pm y,n\in \mathbb{Z}.
So, the correct answer is “Option (c)”.

Note: All possible values of unknown which satisfy the given equation are called solutions of the given equation. For a complete solution “all possible values” satisfying the equation must be obtained. When we try to solve a trigonometric equation, we try to find out all sets of values of the angle, which satisfy the given equation. Sometimes, in simple equations and when it is easy to draw a graph of an equation.