Solutions of Equations of the Form a cos Ө + b sin Ө = c

A trigonometric equation is any equation that contains trigonometric functions. We have learned about trigonometric identities, which are satisfied for every value of the involved angles whereas, trigonometric equations are satisfied only for some values (finite or infinite in number) of the angles. In real life, we use trigonometric equations for making roof inclination, installing ceramic tiles, and building and navigating directions.

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In this article, we will cover the concept of Solutions of Equations of the Form a $\cos \theta+\mathrm{b} \sin \theta=\mathrm{c}$. This category falls under the broader category of Trigonometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), many questions have been asked on this topic.

Trigonometric Equations

Trigonometric equations are, as the name implies, equations that involve trigonometric functions. Trigonometric equations are satisfied only for some values (finite or infinite in number) of the angles. A value of the unknown angle that satisfies the given trigonometric equation is called a solution or a root of the equation. For example, equation $2 \sin x=1$ is satisfied by $x=\pi / 6$ is the solution of the equation between $o$ and $\pi$. The solutions of a trigonometric equation lying in the interval $[0, \pi)$ are called principal solutions.

We know that trigonometric ratios are periodic functions. Functions $\sin x, \cos x, \sec x$, and $\operatorname{cosec} x$ are periodic with period $2 \pi$ and functions $\tan x$ and $\cot x$ are periodic functions with period $\pi$. Therefore, solutions of trigonometric equations can be generalized with the help of the period of trigonometric functions.

Solution of Trigonometric Equation

The value of an unknown angle that satisfies the given trigonometric equation is called a solution or root of the equation. For example, 2sinӨ = 1, clearly Ө = 30⁰ satisfies the equation; therefore, 30⁰ is a solution of the equation. Now trigonometric equation usually has infinite solutions due to the periodic nature of trigonometric functions. So this equation also has (360+30)^o,(720+30)^o,(-360+30)^o, and so on, as its solutions.

Principal Solution

The solutions of a trigonometric equation that lie in the interval [0, 2π). For example, if 2sinӨ = 1, then the two values of sinӨ between 0 and 2π are π/6 and 5π/6. Thus, π/6 and 5π/6 are the principal solutions of equation 2sinӨ = 1.

General Solution

As trigonometric functions are periodic, solutions are repeated within each period, so, trigonometric equations may have an infinite number of solutions. The solution consisting of all possible solutions of a trigonometric equation is called its general solution.

Solutions of Equations of the Form a cos Ө + b sin Ө = c

To solve equations, $a \cos \theta+b \sin \theta=c$, put $a=r \cos \phi$ and $b=r \sin \phi$

Where,

$\\\mathrm{r^2\cos^2\phi+r^2\sin^2\phi=a^2+b^2}\\\mathrm{\Rightarrow \quad r=\sqrt{a^2+b^2}}\\\\\mathrm{ \frac{r\sin\phi}{r\cos\phi}=\tan\phi=\frac{b}{a}}\\\mathrm{\Rightarrow \quad\phi=\tan^{-1}\frac{b}{a}}$

So the equation becomes

$\mathrm{r} \cos \phi \cos \theta+\mathrm{r} \sin \phi \sin \theta=\mathrm{cr}$

$(\cos \phi \cos \theta+\sin \phi \sin \theta)=\cos (\theta-\phi)=\frac{c}{r}$

$\Rightarrow \cos (\theta-\phi)=\frac{\mathrm{c}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}$, if $|\mathrm{c}|>\sqrt{\mathrm{a}^2+\mathrm{b}^2}$, then the equation $\mathrm{a} \cos \theta+\mathrm{b} \sin \theta=\mathrm{c}$ has no solution

Because, when $|\mathrm{c}|>\sqrt{\mathrm{a}^2+\mathrm{b}^2}$ then the value of $\frac{\mathrm{c}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}$ become greater than 1 or less than -1 and range of cos function is $[-1,1]$.

Hence, for a solution to exist for the equation a $\cos \theta+\mathrm{b} \sin \theta=\mathrm{c},|\mathrm{c}|<\sqrt{a^2+b^2}$. In that case, put $\mathrm{\frac{|c|}{\sqrt{a^2+b^2}}=\cos \alpha}$

So our equation becomes,

$\\\mathrm{\cos(\theta-\phi)=\cos \alpha}\\\mathrm{\Rightarrow \quad (\theta-\phi)=2n\pi\pm\alpha}\\\mathrm{\Rightarrow \quad \theta=2n\pi\pm\alpha+\phi,\;\;n\in\mathbb{I}}$

Working Rules for solving such types of equations

1. 1. First of all check whether $|\mathrm{c}| \leq \sqrt{a^2+b^2}$ or not.
2. If $|c|>\sqrt{a^2+b^2}$ then the given equation has no real solution.
3. If $\mid \mathrm{c} \leq \sqrt{a^2+b^2}$, then divide both side of the equation $\sqrt{a^2+b^2}$

Summary

In conclusion, solving a triangle entails utilizing geometric and trigonometric principles to determine the lengths of sides and measures of angles based on given information. To solve such type of equation we need to first determine the equation and should be aware of the general solution of trignometric functions.

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Solved Examples Based on Solutions of Equations of the Form a $\cos \theta+b \sin \theta=c$.

Example 1:If $[\sin x]+[\sqrt{2} \cos x]=-3, x \in[0,2 \pi]_{\text {( }[\text { ] }}$ denotes the greatest integer function) then $x$ belongs to

$
\begin{aligned}
& \text { Solution } \\
& \because \quad[\sin x]+[\sqrt{2} \cos x]=-3 \\
& \Rightarrow \quad[\sin x]=-1 \text { and }[\sqrt{2} \cos x]=-2 \\
& \text { or } \quad-1 \leq \sin x<0 \\
& \text { and } \quad-2 \leq \sqrt{2} \cos x<-1 \\
& \text { or } \quad-1 \leq \sin x<0 \\
& \text { and }-\sqrt{2}<\cos x<-\frac{1}{\sqrt{2}} \\
& \text { Or }-1 \leq \sin x<0 \\
& \text { and }-1 \leq \cos x<-\frac{1}{\sqrt{2}}
\end{aligned}
$

$
\begin{aligned}
& \therefore \quad x \in(\pi, 2 \pi) \text { and } x \in\left(\frac{3 \pi}{4}, \frac{5 \pi}{4}\right) \\
& \therefore \quad x \in(\pi, 2 \pi) \cap\left(\frac{3 \pi}{4}, \frac{5 \pi}{4}\right) \\
& \therefore \quad x \in\left(\pi, \frac{5 \pi}{4}\right) \\
& \text { Hence, the answer is }\left(\pi, \frac{5 \pi}{4}\right)
\end{aligned}
$

Example 2: The general solution of the equation

$
\sum_{r=1}^n \cos \left(r^2 x\right) \sin (r x)=\frac{1}{2}
$
Solution

$
\begin{aligned}
& \because \quad \sum_{r=1}^n \cos \left(r^2 x\right) \sin (r x)=\frac{1}{2} \\
& \Rightarrow \quad \sum_{r=1}^n 2 \cos \left(r^2 x\right) \sin (r x)=1 \\
& \Rightarrow \sum_{r=1}^n[\sin \{r(r+1) x\}-\sin \{r(r-1) x\}]=1
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow \quad \sin \{n(n+1) x\}-\sin 0=1 \\
& \Rightarrow \quad \sin (n(n+1) x)=1 \\
& \therefore \quad n(n+1) x=\left(2 m \pi+\frac{\pi}{2}\right), m \in I \\
& \therefore \quad x=\frac{(4 m+1)}{n(n+1)} \cdot \frac{\pi}{2}, m \in I \\
& \qquad \quad \frac{(4 m+1)}{n(n+1)} \cdot \frac{\pi}{2}, m \in I
\end{aligned}
$

Example 3: Find the general solution of $\sec \theta+\sqrt{3} \tan \theta=1$
Solution

$
\begin{aligned}
& \sec \theta+\sqrt{3} \tan \theta=1 \\
& \frac{1}{\cos \theta}+\sqrt{3} \frac{\sin \theta}{\cos \theta}=1 \\
& \sqrt{3} \sin \theta-\cos \theta=-1 \\
& a=\sqrt{3}, b=-1, \sqrt{a^2+b^2}=\sqrt{3+1}=2
\end{aligned}
$

Dividing the equation by 2

$
\begin{aligned}
& -\cos \left(\theta+\frac{\pi}{3}\right)=-\frac{1}{2} \\
& \cos \left(\theta+\frac{\pi}{3}\right)=\cos \frac{\pi}{3} \\
& \theta+\frac{\pi}{3}=2 n \pi \pm \frac{\pi}{3} \\
& \theta=2 n \pi \pm \frac{\pi}{3}-\frac{\pi}{3}
\end{aligned}
$

Hence, the answer is

$
2 n \pi \pm \frac{\pi}{3}-\frac{\pi}{3}
$

Example 4: The general solution of the equation $\sqrt{3} \sin x+\cos x=1$ is:
Solution

$
\begin{aligned}
& \sqrt{3} \sin x+\cos x=1 \\
& a=\sqrt{3}, b=1, \sqrt{a^2+b^2}=\sqrt{3+1}=2
\end{aligned}
$
Dividing the equation by 2

$
\begin{aligned}
& \frac{\sqrt{3}}{2} \sin x+\frac{1}{2} \cos x=\frac{1}{2} \\
& \sin \frac{\pi}{3} \sin x+\cos \frac{\pi}{3} \cos x=\frac{1}{2} \\
& \cos \left(x-\frac{\pi}{3}\right)=\cos \frac{\pi}{3} \\
& x-\frac{\pi}{3}=2 n \pi \pm \frac{\pi}{3} \\
& x=2 n \pi \pm \frac{\pi}{3}+\frac{\pi}{3}
\end{aligned}
$

Hence, the answer is

$
2 n \pi \pm \frac{\pi}{3}+\frac{\pi}{3}
$

Example 5: What is the Range of $t$ , where $t \epsilon[-\pi, \pi]$ and $\sin x+\sin (t+x)+\sin (t-x)=1$ has real roots?

Solution

Example 5: What is the Range of $\$ \mathrm{t} \$$, where $\$ \mathrm{t}$ lepsilon[-lpi, 1 pi$] \$$ and $\$ 1 \mathrm{sin}$ $x+\backslash \sin (t+x)+\backslash \sin (t-x)=1 \$$ has real roots?|

Solution

$
\begin{aligned}
& \sin x+\sin (t+x)+\sin (t-x)=1 \\
& \sin x+2 \sin t \cdot \cos x=1
\end{aligned}
$

Here $\mathrm{a}=2 \sin \mathrm{t}, \mathrm{b}=1$ and $\mathrm{c}=1$
For real roots $|c| \leq \sqrt{a^2+b^2}$

$
\begin{aligned}
& 1 \leq \sqrt{1+2 \sin t} \\
& 1 \leq 1+2 \sin t \\
& \sin t \geq 0 \\
& t \in[0, \pi]
\end{aligned}
$
Hence, the answer is $[0, \pi]$