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
300.6k+ 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.
Recently Updated Pages
If a parabola whose length of latus rectum is 4a touches class 11 maths JEE_Main

Find the cubic polynomial whose zeroes are 3 5 and class 11 maths JEE_Main

During the sale colour pencils were being sold in -class-11-maths-JEE_Main

A man on the top of a vertical observation tower o-class-11-maths-JEE_Main

In a class of 60 students 25 students play cricket class 11 maths JEE_Main

A regular polygon has 20 sides How many triangles can class 11 maths JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
NCERT Solutions For Class 11 Maths Chapter 6 Permutations And Combinations - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Maths Chapter 9 Straight Lines - 2026-27 Free PDF Download (Sign-in Required)

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

NCERT Solutions For Class 11 Maths Chapter 8 Sequences And Series - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Maths Chapter 4 Complex Numbers And Quadratic Equations - 2026-27 Free PDF Download (Login Required)

Hybridisation in Chemistry – Concept, Types & Applications

