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Question: The general solution of equation \(\sin 2x + 2\sin x + 2\cos x + 1 = 0\) A. \(3n\pi - \dfrac{\pi ...

The general solution of equation sin⁡2x+2sin⁡x+2cos⁡x+1=0\sin 2x + 2\sin x + 2\cos x + 1 = 0
A. 3nπ−π43n\pi - \dfrac{\pi }{4}
B. 2nπ+π42n\pi + \dfrac{\pi }{4}
C. 2nπ+(−1)nsin⁡−1(13)2n\pi + {\left( { - 1} \right)^n}{\sin ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right)
D. nπ−π4n\pi - \dfrac{\pi }{4}

Explanation

Solution

Hint: We have to calculate the given equation by using different trigonometric formulas and solve it. After that by using these formulas change the format of the equation and write it in the form which we want to get it by observing the given options. Thus we can solve the problem very easily.

Complete step-by-step answer:
First we proceed with the given equation is-
sin⁡2x+2sin⁡x+2cos⁡x+1=0\sin 2x + 2\sin x + 2\cos x + 1 = 0 -------(1)\left( 1 \right)
We know that from the trigonometry formula that-
sin⁡2x+cos⁡2x=1 And sin⁡2x= 2sin⁡xcos⁡x  {\sin ^2}x + {\cos ^2}x = 1 \\\ {\text{And }}\sin 2x = {\text{ }}2\sin x\cos x \\\
Now, we write the equation in different form, i.e
1+sin⁡2x+2(sin⁡x+cos⁡x)=01 + \sin 2x + 2\left( {\sin x + \cos x} \right) = 0
We have taken 2 common in the above equation
Now by applying formulas in equation (1)\left( 1 \right)
We get-
sin⁡2x+cos⁡2x+2sin⁡xcos⁡x+2(sin⁡x+cos⁡x)=0 ⇒(sin⁡x+cos⁡x)2+2(sin⁡x+cos⁡x)=0  {\sin ^2}x + {\cos ^2}x + 2\sin x\cos x + 2\left( {\sin x + \cos x} \right) = 0 \\\ \Rightarrow {\left( {\sin x + \cos x} \right)^2} + 2\left( {\sin x + \cos x} \right) = 0 \\\
This is done by using the formula of a2+b2+2ab{a^2} + {b^2} + 2ab in the first three terms of the equation.
Because we know that (a+b)2=a2+b2+2ab{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab
Now by taking sin⁡x,cos⁡x\sin x,\cos x in the solved equation
We get
(sin⁡x+cos⁡x)(sin⁡x+cos⁡x+2)=0  \left( {\sin x + \cos x} \right)\left( {\sin x + \cos x + 2} \right) = 0 \\\
From this, we get
sin⁡x+cos⁡x=0 or sin⁡x+cos⁡x=−2  \sin x + \cos x = 0{\text{ or}} \\\ \sin x + \cos x = - 2 \\\
But sin⁡x+cos⁡x=−2\sin x + \cos x = - 2 is not inadmissible because of negative value.
We know that ∣sin⁡x∣⩽1,∣cos⁡x∣⩽1\left| {\sin x} \right| \leqslant 1,\left| {\cos x} \right| \leqslant 1
So it can never be -2
Therefore sin⁡x+cos⁡x=0\sin x + \cos x = 0
It can be written as sin⁡(x+π4)=0\sin \left( {x + \dfrac{\pi }{4}} \right) = 0
⇒x+π4=nπ ⇒x=nπ−π4  \Rightarrow x + \dfrac{\pi }{4} = n\pi \\\ \Rightarrow x = n\pi - \dfrac{\pi }{4} \\\
Since we get the right answer
Hence option D is correct.

Note- We have the trigonometric formulas to crack the equation in desired format and after that solve it by putting those formulas. Since we get the equation in desired format and get the value of xx. We have to solve the problem with the help of given options.