Simultaneous Trigonometric Equations

The trigonometric equations are the equations that contain functions of the trigonometry. 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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Simultaneous 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.

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, $2 \sin \Theta=1$, clearly $\Theta=30^0$ satisfies the equation; therefore, $30^{\circ}$ 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 $(360+30)^0,(720+30)^0,(-360+30)^0$ and so on, as its solutions.

Principal Solution
The solutions of a trigonometric equation that lie in the interval $[0,2 \pi)$. For example, if $2 \sin \Theta=1$, then the two values of $\sin \Theta$ between 0 and $2 \pi$ are $\pi / 6$ and $5 \pi / 6$. Thus, $\pi / 6$ and $5 \pi / 6$ are the principal solutions of equation $2 \sin \theta=$ 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.

Sometimes the value of angle(s) that satisfies more than one trigonometric equation simultaneously is asked. We call such a system of equations as Simultaneous Trigonometric Equations

Illustration

General value of $x$ which satisfies the equation $\cos x=-1 / 2$ and $\cot x=1 / \sqrt{ } 3$

First, find the value of $x$ in $[0,2 \pi)$ which satisfies two given equations separately
$
\begin{aligned}
& \cos x=-1 / 2 \Rightarrow x=2 \pi / 3 \text { and } 4 \pi / 3 \\
& \cot x=1 / \sqrt{ } 3 \Rightarrow x=\pi / 3 \text { and } 4 \pi / 3
\end{aligned}
$

Now select the value $x$ which satisfies both the equation
Here common value is $4 \pi / 3$
Also, we know that cos and cos functions will repeat after an interval of $2 \pi$, so the required general solution is $x=2 n \pi+4 \pi / 3$

Summary: Solving trigonometric equations is an important concept in mathematics. By applying these formulas, complex equations can be solved easily with the help of these formulas. Trigonometric Equation help analysts to provide insights based on real-life based on Science, commerce, etc.

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Solved Examples Based on Trigonometric Equations:

Example 1: Find the smallest positive root of the

$
\sqrt{\cos (1-x)}=\sqrt{\sin x}
$
1) $\frac{\pi}{4}-\frac{1}{2}$
2) $\frac{\pi}{4}+\frac{1}{2}$
3) $\frac{\pi}{2}-\frac{1}{2}$
4) None of these

Solution

$
\begin{aligned}
& \quad \sqrt{\cos (1-x)}=\sqrt{\sin x} \\
& \cos (1-x) \geq 0 \text { and } \sin x \geq 0 \\
& \cos (1-x)=\sin x \\
& \sin \left(\frac{\pi}{2}-(1-x)\right)=\sin x \\
& \frac{\pi}{2}-1+x=n \pi+(-1)^n x \\
& \text { at } n=1 \\
& 2 x=\frac{\pi}{2}+1
\end{aligned}
$
$
x=\frac{\pi^{-}}{4}+\frac{1}{2}
$

For this value of x both satisfies $\cos (1-x) \geq 0$ and $\sin x \geq 0$

Hence, the answer is option (2).

Example 2: Find the number of solutions for $\cos x=\frac{x}{5}$
1) 1
2) 2
3) 3
4) 4

Key Concepts
Eliminating $x$ from above equation, we get

$
\begin{aligned}
& \mathrm{r}=3 \cdot \frac{4}{\mathrm{r}}-1 \\
& \Rightarrow \mathrm{r}^2=12-\mathrm{r} \\
& \Rightarrow \mathrm{r}=3,-4
\end{aligned}
$

Now,

$
r \sin x=4 \Rightarrow \sin x=\frac{4}{-4}=-1
$

and, $\sin x=\frac{4}{3}$ which is not possible solve $\sin \mathrm{x}=-1$

$
r \sin x=4 \Rightarrow \sin x=\frac{4}{-4}=-1
$

and, $\sin \mathrm{x}=\frac{4}{3}$ which is not possible solve $\sin x=-1$

$
\Rightarrow \mathrm{x}=-\frac{n}{2}
$

Hence the required pair is ( $-4,-\pi / 2$ )

$
\begin{aligned}
& \quad \cos x=\frac{x}{5} \\
& -1 \leq \cos x \leq 1 \\
& -5 \leq x \leq 5 \\
& \text { at } \mathrm{x}=2 \pi \\
& \frac{x}{5}>1
\end{aligned}
$

Graph -

By the graph we can say the number of solutions for this equation is 3.

Example 3: Find the number of solution/solutions for $(\cos x+\sec x)^2=4, x \epsilon[0, \pi]$
1) $0$
2) $1$
3) $2$
4) $3$

Solution
$(\cos x+\sec x)^2=2$ asssume $\cos x=\mathrm{t}\left(t+\frac{1}{t}\right)^2=t^2+\left(\frac{1}{t}\right)^2+2 \geq 2$ L.H.S. $\geq 2$ R.H.S. $=2$ L.H.S. and R.H.S. is same if $\cos x=\sec x x=0$
Hence, the answer is the option 2.

Example 4: Let $A=\{\theta: \sin (\theta)=\tan (\theta)\}$ and $B=\{\theta: \cos (\theta)=1\}$ be two sets. Then:
1) $A=B$
2) $A \nsubseteq B$
3) $B \nsubseteq A$
4) $A \subset B$ and $B-A \neq \phi$

Solution
Given,

$
\begin{aligned}
& A=\{\theta: \sin \theta=\tan \theta\} \\
& B=\{\theta: \cos \theta=1\}
\end{aligned}
$

Now,

$
\begin{aligned}
& A=\left\{\theta: \sin \theta=\frac{\sin \theta}{\cos \theta}\right\} \\
& =\{\theta: \sin \theta(\cos \theta-1)=0\} \\
& =\{\theta=0, \pi, 2 \pi, 3 \pi, \ldots\}
\end{aligned}
$
For $B: \cos \theta=1 \Rightarrow \theta=\pi, 2 \pi, 4 \pi, \ldots \ldots$.
This shows that $A$ is not contained in B . i.e. $A \not \subset B$. but $B \subset A$

Hence, the answer is option 2.

Example 5: Statement 1: The number of common solutions of the trigonometric equations $2 \sin ^2 \theta-\cos 2 \theta=0$ and $2 \cos ^2 \theta-3 \sin \theta=0$ in the interval $[0,2 \pi]$ is two :

Statement 2: The number of solutions of the equation, $2 \cos ^2 \theta-3 \sin \theta=0$ in the interval $[0, \pi]$ is two.

1) Statement 1 is true; Statement 2 is true ; Statement 2 is a correct explanation for Statement 1

2) Statement 1 is true; Statement 2 is true; Statement 2 is not a correct explanation for Statement 1

3) Statement 1 is false; Statement 2 is true.

4) Statement 1 is true; Statement 2 is false.

Solution
$
\begin{aligned}
& 2 \sin ^2 \theta-\cos 2 \theta=0 \\
& \Rightarrow 2 \sin ^2 \theta-\left(1-2 \sin ^2 \theta\right)=0 \\
& \Rightarrow 2 \sin ^2 \theta-1+2 \sin ^2 \theta=0 \\
& \Rightarrow 4 \sin ^2 \theta=1 \Rightarrow \sin \theta= \pm \frac{1}{2} \\
& \therefore \theta=\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{7 \pi}{6}, \frac{11 \pi}{6}
\end{aligned}
$

Now $2 \cos ^2 \theta-3 \sin \theta=0$

$
\begin{aligned}
& \Rightarrow 2\left(1-\sin ^2 \theta\right)-3 \sin \theta=0 \\
& \Rightarrow-2 \sin ^2 \theta-3 \sin \theta+2=0 \\
& \Rightarrow-2 \sin ^2 \theta-4 \sin \theta+\sin \theta+2=0 \\
& \Rightarrow 2 \sin ^2 \theta-\sin \theta+4 \sin \theta-2=0 \\
& \Rightarrow \sin \theta(2 \sin \theta-1)+2(2 \sin \theta-1)=0 \\
& \Rightarrow \sin \theta=\frac{1}{2},-2
\end{aligned}
$
But $\sin \theta=-2$, is not possible

$
\therefore \quad \sin \theta=\frac{1}{2}, \quad \Rightarrow \quad \theta=\frac{\pi}{6}, \frac{5 \pi}{6}
$

Hence, there are two common solutions, where each of the statements- 1 and 2 is true but statement- 2 is not a correct explanation for statement 1.

Hence, the answer is the option (2).