
If $4{{\sin }^{2}}\theta +2(\sqrt{3}+1)\cos \theta =4+\sqrt{3}$, then the general value of$\theta $ is
A. \[2n\pi \pm \dfrac{\pi }{3}\]
B. \[2n\pi +\dfrac{\pi }{4}\]
C. \[n\pi \pm \dfrac{\pi }{3}\]
D. \[n\pi -\dfrac{\pi }{3}\]
Answer
163.2k+ views
Hint: To find the general value of $\theta $, we will consider the given equation and use the formula of ${{\sin }^{2}}\theta $ in it and simplify the equation. Then we will get a resultant quadratic equation which we will factorize and derive two factors. We will then use the theorem which states that if $x$ and $y$are real numbers, then $\cos x=\cos y$implies that $x=2n\pi \pm y$, where $n\in Z$ and is an integer.
Complete step by step solution:We are given $4{{\sin }^{2}}\theta +2(\sqrt{3}+1)\cos \theta =4+\sqrt{3}$ and we have to determine the general value of $\theta $.
We will take the given equation and use the formula of ${{\sin }^{2}}\theta $ in it and simplify.
\[\begin{align}
& 4{{\sin }^{2}}\theta +2(\sqrt{3}+1)\cos \theta =4+\sqrt{3} \\
& 4(1-{{\cos }^{2}}\theta )+2(\sqrt{3}+1)\cos \theta =4+\sqrt{3} \\
& 4-4{{\cos }^{2}}\theta +2(\sqrt{3}+1)\cos \theta =4+\sqrt{3} \\
& -4{{\cos }^{2}}\theta +2(\sqrt{3}+1)\cos \theta =\sqrt{3} \\
& 4{{\cos }^{2}}\theta -2(\sqrt{3}+1)\cos \theta +\sqrt{3}=0
\end{align}\]
We can see that the resultant equation is a quadratic equation so we will now factorize this equation and derive the factors.
\[\begin{align}
& 4{{\cos }^{2}}\theta -2(\sqrt{3}+1)\cos \theta +\sqrt{3}=0 \\
& 4{{\cos }^{2}}\theta -2\sqrt{3}\cos \theta -2\cos \theta +\sqrt{3}=0 \\
& 2\cos \theta (2\cos \theta -\sqrt{3})-1(2\cos \theta -\sqrt{3})=0 \\
& (2\cos \theta -\sqrt{3})(2\cos \theta -1)=0
\end{align}\]
Now we will equate both the factors to zero.
\[\begin{align}
& 2\cos \theta -\sqrt{3}=0 \\
& 2\cos \theta =\sqrt{3} \\
& \cos \theta =\dfrac{\sqrt{3}}{2} \\
& \cos \theta =\cos \dfrac{\pi }{6}
\end{align}\] Or \[\begin{align}
& 2\cos \theta -1=0 \\
& \cos \theta =\dfrac{1}{2} \\
& \cos \theta =\cos \dfrac{\pi }{6}
\end{align}\]
Applying the theorem for cosine here we will derive the general value of $\theta $.
$\theta =2n\pi \pm \dfrac{\pi }{6}$ Or $\theta =2n\pi \pm \dfrac{\pi }{3}$
Here $n\in Z$ and is an integer.
In the options given, the general value of $\theta $ is $\theta =2n\pi \pm \dfrac{\pi }{3}$ therefore this will be our correct option.
The general value of $\theta $ is $\theta =2n\pi \pm \dfrac{\pi }{3}$ when $4{{\sin }^{2}}\theta +2(\sqrt{3}+1)\cos \theta =4+\sqrt{3}$.
Option ‘A’ is correct
Note: We should know all the theorems for all the trigonometric functions to derive the general value of any angle. The general value of angle also helps in deriving the principal solutions for any question because by substituting the value of $n$, we can get all the possible solutions. We should also know all the formulas of trigonometric functions and their table of values for each angle.
Complete step by step solution:We are given $4{{\sin }^{2}}\theta +2(\sqrt{3}+1)\cos \theta =4+\sqrt{3}$ and we have to determine the general value of $\theta $.
We will take the given equation and use the formula of ${{\sin }^{2}}\theta $ in it and simplify.
\[\begin{align}
& 4{{\sin }^{2}}\theta +2(\sqrt{3}+1)\cos \theta =4+\sqrt{3} \\
& 4(1-{{\cos }^{2}}\theta )+2(\sqrt{3}+1)\cos \theta =4+\sqrt{3} \\
& 4-4{{\cos }^{2}}\theta +2(\sqrt{3}+1)\cos \theta =4+\sqrt{3} \\
& -4{{\cos }^{2}}\theta +2(\sqrt{3}+1)\cos \theta =\sqrt{3} \\
& 4{{\cos }^{2}}\theta -2(\sqrt{3}+1)\cos \theta +\sqrt{3}=0
\end{align}\]
We can see that the resultant equation is a quadratic equation so we will now factorize this equation and derive the factors.
\[\begin{align}
& 4{{\cos }^{2}}\theta -2(\sqrt{3}+1)\cos \theta +\sqrt{3}=0 \\
& 4{{\cos }^{2}}\theta -2\sqrt{3}\cos \theta -2\cos \theta +\sqrt{3}=0 \\
& 2\cos \theta (2\cos \theta -\sqrt{3})-1(2\cos \theta -\sqrt{3})=0 \\
& (2\cos \theta -\sqrt{3})(2\cos \theta -1)=0
\end{align}\]
Now we will equate both the factors to zero.
\[\begin{align}
& 2\cos \theta -\sqrt{3}=0 \\
& 2\cos \theta =\sqrt{3} \\
& \cos \theta =\dfrac{\sqrt{3}}{2} \\
& \cos \theta =\cos \dfrac{\pi }{6}
\end{align}\] Or \[\begin{align}
& 2\cos \theta -1=0 \\
& \cos \theta =\dfrac{1}{2} \\
& \cos \theta =\cos \dfrac{\pi }{6}
\end{align}\]
Applying the theorem for cosine here we will derive the general value of $\theta $.
$\theta =2n\pi \pm \dfrac{\pi }{6}$ Or $\theta =2n\pi \pm \dfrac{\pi }{3}$
Here $n\in Z$ and is an integer.
In the options given, the general value of $\theta $ is $\theta =2n\pi \pm \dfrac{\pi }{3}$ therefore this will be our correct option.
The general value of $\theta $ is $\theta =2n\pi \pm \dfrac{\pi }{3}$ when $4{{\sin }^{2}}\theta +2(\sqrt{3}+1)\cos \theta =4+\sqrt{3}$.
Option ‘A’ is correct
Note: We should know all the theorems for all the trigonometric functions to derive the general value of any angle. The general value of angle also helps in deriving the principal solutions for any question because by substituting the value of $n$, we can get all the possible solutions. We should also know all the formulas of trigonometric functions and their table of values for each angle.
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