Solving simultaneous trigonometric equations often requires combining identities, factorization, and substitution to handle equations involving $\sin \theta$, $\cos \theta$, or other trigonometric functions. These equations frequently appear in higher-level problems where multiple trigonometric conditions need to be satisfied together. Mastering them is useful not only for Class 11 but also for tackling advanced math and competitive exam questions. In this article, we cover simultaneous trigonometric equations examples, different solving techniques, and step-by-step methods on how to solve a system of equations with trigonometric functions in mathematics.
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Trigonometric equations are equations that involve trigonometric functions like $\sin \theta$, $\cos \theta$, or $\tan \theta$. These equations are satisfied only for specific values (finite or infinite) of the angle. A value of the unknown angle that satisfies the given trigonometric equation is called a solution or a root of the equation.
For example, the equation $2 \sin x = 1$ is satisfied by $x = \pi/6$ within the interval $[0, \pi)$. Such values are called principal solutions.
A solution or root of the equation is the value of the unknown angle that makes the equation true.
Example: $2 \sin \theta = 1$ gives $\theta = 30^\circ$ as a solution.
Since trigonometric functions are periodic, equations usually have infinitely many solutions.
Thus, solutions also include $(360^\circ + 30^\circ), (720^\circ + 30^\circ), (-360^\circ + 30^\circ)$, and so on.
The solutions of a trigonometric equation that lie in the interval $[0, 2\pi)$ are called principal solutions.
Example: If $2 \sin \theta = 1$, then the values of $\theta$ between $0$ and $2\pi$ are $\pi/6$ and $5\pi/6$.
Hence, $\pi/6$ and $5\pi/6$ are the principal solutions.
Because trigonometric functions are periodic, solutions repeat after each period.
The general solution includes all possible solutions of the equation.
For example, if $\sin \theta = \tfrac{1}{2}$, then the general solution is:
$\theta = n\pi + (-1)^n \tfrac{\pi}{6}, n \in \mathbb{Z}$
Sometimes, the value of $\theta$ must satisfy more than one trigonometric equation simultaneously. Such a system is called a simultaneous trigonometric equation.
Find the general solution of $x$ which satisfies both:
$\cos x = -\tfrac{1}{2}$
$\cot x = \tfrac{1}{\sqrt{3}}$
Step 1: Solve separately in $[0, 2\pi)$
$\cos x = -\tfrac{1}{2} \Rightarrow x = \tfrac{2\pi}{3}, \tfrac{4\pi}{3}$
$\cot x = \tfrac{1}{\sqrt{3}} \Rightarrow x = \tfrac{\pi}{3}, \tfrac{4\pi}{3}$
Step 2: Find the common solution
The common value is $x = \tfrac{4\pi}{3}$.
Step 3: Write the general solution
Since trigonometric functions repeat every $2\pi$,
$x = 2n\pi + \tfrac{4\pi}{3}, n \in \mathbb{Z}$
Below are the standard methods to solve simultaneous trigonometric equations, covering approaches like trigonometric identities, substitution of $\sin \theta$ and $\cos \theta$, factorization techniques, and elimination strategies for handling systems of trigonometric equations effectively.
Apply fundamental trigonometric identities like $\sin^2 \theta + \cos^2 \theta = 1$, $1 + \tan^2 \theta = \sec^2 \theta$, etc. Helps in reducing equations with multiple functions into a single function.
Example: If $\sin \theta + \cos \theta = \sqrt{2}$, squaring both sides and using $\sin^2 \theta + \cos^2 \theta = 1$ gives a solvable form.
Often useful when equations involve both $\sin$ and $\cos$. Converts complex systems into simpler, solvable equations.
Replace $\sin \theta = p$ and $\cos \theta = q$. Use the condition $p^2 + q^2 = 1$ to validate solutions. Reduces the system into linear or quadratic equations in $p$ and $q$. Helps avoid directly dealing with trigonometric complexity. Example: If $\sin \theta + \cos \theta = 1$, substituting $p$ and $q$ makes it a linear equation with constraint $p^2 + q^2 = 1$.
Factorize expressions like polynomials.
Example: $\sin^2 \theta - \sin \theta = 0 \implies \sin \theta (\sin \theta - 1) = 0$.
Gives multiple solutions at once. Best applied when equations are quadratic or higher order in $\sin$ or $\cos$. Works effectively in simultaneous equations where one equation is factorable.
Similar to elimination in algebraic systems. Multiply or rearrange equations to cancel out one trigonometric term.
Example: If $\sin \theta + \cos \theta = 1$ and $\sin \theta - \cos \theta = 0$, adding them eliminates $\cos \theta$.
Reduces two equations into one trigonometric function. Works well when both equations involve the same set of trigonometric ratios.
Below are the examples, including problems with $\sin \theta$ and $\cos \theta$, equations involving $\tan \theta$, $\cot \theta$, $\sec \theta$, and $\csc \theta$, as well as word problems to strengthen problem-solving skills.
Example: Solve $\sin \theta + \cos \theta = 1$ and $\sin \theta - \cos \theta = 0$.
Adding gives $2 \sin \theta = 1 \implies \sin \theta = 1/2 \implies \theta = \pi/6, 5\pi/6$. Substitute into second equation to find valid $\theta$. Demonstrates use of elimination and substitution.
Convert these into $\sin$ and $\cos$ for easier handling.
Example: Solve $\tan \theta = \sqrt{3}$ and $\sec \theta = 2$.
From $\tan \theta = \sqrt{3} \implies \theta = \pi/3, 4\pi/3$. Check with $\sec \theta = 2 \implies \cos \theta = 1/2 \implies \theta = \pi/3, 5\pi/3$. Common solution: $\theta = \pi/3$.
Below are the key points on the graphical method for solving simultaneous trigonometric equations, where solutions are obtained by observing the intersection of graphs of $\sin \theta$, $\cos \theta$, $\tan \theta$ and other trigonometric functions for common values of $\theta$.
Plot $\sin \theta$ and $\cos \theta$ in the same interval.
Intersection points correspond to solutions.
Example: $\sin \theta = \cos \theta$ gives intersections at $\theta = \pi/4, 5\pi/4$.
Useful for quick visualization of solutions.
For equations like $\sin \theta + \cos \theta = 1$, graph $y = \sin \theta + \cos \theta$ and $y = 1$. Solutions are where curves intersect.
Helps verify number of solutions within $[0,2\pi)$. Provides intuitive understanding beyond algebraic manipulation.
Below are some advanced techniques and tricks for solving simultaneous trigonometric equations, including the use of the auxiliary angle method, polynomial transformation, and other shortcuts that simplify complex trigonometric systems into solvable forms.
Convert expressions like $a \sin \theta + b \cos \theta$ into $R \sin (\theta + \alpha)$. Simplifies solving simultaneous equations.
Example: Solve $3 \sin \theta + 4 \cos \theta = 5$. Write as $5 \sin (\theta + \alpha)$ and solve directly. Avoids long algebraic manipulation.
Use identities such as $\tan(\theta/2) = t$ to transform equations. Converts trigonometric equations into polynomial equations in $t$.
Example: $\sin \theta = \cos \theta$ becomes $\frac{2t}{1+t^2} = \frac{1-t^2}{1+t^2}$.
Solve quadratic in $t$, then back-substitute for $\theta$. Particularly useful for higher-order or complex systems.
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:
$\sqrt{\cos (1-x)}=\sqrt{\sin x}$
$\cos (1-x) \geq 0 \text { and } \sin x \geq 0$
$\cos (1-x)=\sin x$
$\sin \left(\frac{\pi}{2}-(1-x)\right)=\sin x$
$\frac{\pi}{2}-1+x=n \pi+(-1)^n x$
$\text { at } n=1$
$2 x=\frac{\pi}{2}+1$
$x=\frac{\pi^{-}}{4}+\frac{1}{2}$
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
Solution:
Eliminating $x$ from above equation, we get
$r = 3 \cdot \tfrac{4}{r} - 1$
$\Rightarrow r^2 = 12 - r$
$\Rightarrow r = 3, -4$
Now, $r \sin x = 4 \Rightarrow \sin x = \tfrac{4}{-4} = -1$
and $\sin x = \tfrac{4}{3}$ which is not possible.
So we solve $\sin x = -1$
$\Rightarrow x = -\tfrac{\pi}{2}$
Hence, the required pair is $(-4, -\pi/2)$
Next, $\cos x = \tfrac{x}{5}$
$-1 \leq \cos x \leq 1$
$\Rightarrow -5 \leq x \leq 5$
At $x = 2\pi$,
$\tfrac{x}{5} > 1$
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, $A = \{\theta : \sin \theta = \tan \theta\}$
$B = \{\theta : \cos \theta = 1\}$
Now, $A = \left\{ \theta : \sin \theta = \tfrac{\sin \theta}{\cos \theta} \right\}$
$A = \{\theta : \sin \theta (\cos \theta - 1) = 0\}$
$A = \{\theta = 0, \pi, 2\pi, 3\pi, \ldots\}$
For $B$: $\cos \theta = 1 \Rightarrow \theta = 2n\pi, \; n \in \mathbb{Z}$
This shows that $A \not\subset B$, but $B \subset A$.
Hence, the answer is Option (2).
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:
$2 \sin^2 \theta - \cos 2\theta = 0$
$2 \sin^2 \theta - (1 - 2 \sin^2 \theta) = 0$
$2 \sin^2 \theta - 1 + 2 \sin^2 \theta = 0$
$4 \sin^2 \theta = 1$
$\sin \theta = \pm \tfrac{1}{2}$
$\therefore \theta = \tfrac{\pi}{6}, \tfrac{5\pi}{6}, \tfrac{7\pi}{6}, \tfrac{11\pi}{6}$
Now, solve
$2 \cos^2 \theta - 3 \sin \theta = 0$
$2(1 - \sin^2 \theta) - 3 \sin \theta = 0$
$-2 \sin^2 \theta - 3 \sin \theta + 2 = 0$
$-2 \sin^2 \theta - 4 \sin \theta + \sin \theta + 2 = 0$
$(2 \sin^2 \theta - \sin \theta) + (4 \sin \theta - 2) = 0$
$\sin \theta (2 \sin \theta - 1) + 2(2 \sin \theta - 1) = 0$
$(2 \sin \theta - 1)(\sin \theta + 2) = 0$
$\sin \theta = \tfrac{1}{2}, -2$
But $\sin \theta = -2$ is not possible.
$\therefore \sin \theta = \tfrac{1}{2}$
$\theta = \tfrac{\pi}{6}, \tfrac{5\pi}{6}$
Hence, there are two common solutions. Both statements 1 and 2 are true, but statement 2 is not the correct explanation of statement 1.
Answer: Option (2)
Below is a list of important topics related to simultaneous trigonometric equations, covering methods, examples, and applications useful for Class 11 and competitive exams.
Below are the NCERT resources on Trigonometric Functions, including detailed notes, solutions, and exemplar problems for proper learning and preparation of competitive exams.
NCERT Class 11 Chapter 3 - Trigonometric Functions Notes
NCERT Class 11 solutions for Chapter 3 - Trigonometric Functions
NCERT Exemplar solutions for Class 11 Chapter 3 - Trigonometric Functions
Below are carefully selected practice questions on simultaneous trigonometric equations, covering problems with $\sin \theta$, $\cos \theta$, $\tan \theta$, and other functions. These examples will help strengthen problem-solving skills and build accuracy in handling systems of trigonometric equations.
Simultaneous Trigonometric Equations - Practice Question
We have provided below the links to practice questions on the related topics:
Frequently Asked Questions (FAQs)
The 7 basic trigonometry formulas include ratios: $\sin \theta = \frac{opposite}{hypotenuse}$, $\cos \theta = \frac{adjacent}{hypotenuse}$, $\tan \theta = \frac{opposite}{adjacent}$, and their reciprocals $\csc \theta$, $\sec \theta$, $\cot \theta$, along with the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$.
The four common methods are substitution, elimination, cross-multiplication, and graphical method. These can also be adapted for simultaneous trigonometric equations involving $\sin \theta$ and $\cos \theta$.
A simultaneous equation is a system where two or more equations are solved together to find common values of unknowns. In trigonometry, it often involves $\sin \theta$, $\cos \theta$, or $\tan \theta$ solved within the same system.
Simultaneous quadratic equations are solved by substitution, elimination, or factorization. In trigonometric form, quadratic equations like $a \sin^2 \theta + b \sin \theta + c = 0$ can be reduced to standard quadratic form and solved accordingly.
Trigonometric equations are classified as linear trigonometric equations, quadratic trigonometric equations, multiple-angle equations, and simultaneous trigonometric equations, depending on the form and degree of trigonometric functions involved.