Find the principal solutions of each of the following equations:
(i) $\sin x=\frac{\sqrt{3}}{2}$
(ii) $\cos \mathrm{x}=\frac{1}{2}$
(iii) $\tan \mathrm{x}=\sqrt{3}$
(iv) $\cot \mathrm{x}=\sqrt{3}$
(v) $\operatorname{cosec} x=2$
(vi) $\sec x=\frac{2}{\sqrt{3}}$
To Find: Principal solution.
[NOTE: The solutions of a trigonometry equation for which 0 ≤ x < 2 ???? is called principal solution]
(i) Given: $\sin x=\frac{\sqrt{3}}{2}$
Formula used: $\sin \theta=\sin \alpha \Rightarrow \theta=\mathrm{n} \pi+(-1)^{\mathrm{n}} \alpha, \mathrm{n} \in \mathrm{I}$
By using above formula, we have
$\sin \mathrm{x}=\frac{\sqrt{3}}{2}=\sin \frac{\pi}{3} \Rightarrow \mathrm{x}=\mathrm{n} \pi+\frac{\pi}{3}(-1)^{\mathrm{n}}$
Put $n=0 \Rightarrow x=0 \times \pi+\frac{\pi}{3}(-1)^{0} \Rightarrow x=\frac{\pi}{3}$
Put $n=1 \Rightarrow x=1 \times \pi+\frac{\pi}{3}(-1)^{1} \Rightarrow x=1 \times \pi+\frac{\pi}{3}(-1)^{1} \Rightarrow x=\pi-\frac{\pi}{3}=\frac{2 \pi}{3}$
So principal solution is $x=\frac{\pi}{3}$ and $\frac{2 \pi}{3}$
(ii) Given: $\cos x=\frac{1}{2}$
Formula used: $\cos \theta=\cos \alpha \Rightarrow \theta=2 n \pi \pm \alpha, n \in I$
By using above formula, we have
$\cos x=\frac{1}{2}=\cos \frac{\pi}{3} \Rightarrow \theta=2 n \pi \pm \alpha, n \in 1$
Put $n=0 \Rightarrow x=2 n \pi \pm \frac{\pi}{3} \Rightarrow x=\frac{\pi}{3}$
Put $n=1 \Longrightarrow x=2 \pi \pm \frac{\pi}{3} \Rightarrow x=\frac{5 \pi}{3}, \frac{7 \pi}{3} \Rightarrow x=\frac{5 \pi}{3}, \frac{7 \pi}{3}$
$\left[\frac{7 \pi}{3}>2 \pi\right.$ So it is not include in principal solution $]$
So principal solution is $x=\frac{\pi}{3}$ and $\frac{5 \pi}{3}$
(iii) Given: $\tan \mathrm{x}=\sqrt{3}$
Formula used: $\tan \theta=\tan \alpha \Rightarrow \theta=n \pi \pm \alpha, n \in \mid$
By using above formula, we have
$\tan \mathrm{x}=\sqrt{3}=\tan \frac{\pi}{3} \Rightarrow \mathrm{x}=\mathrm{n} \pi+\alpha, \mathrm{n} \in 1$
Put $\mathrm{n}=0 \Rightarrow \mathrm{x}=\mathrm{n} \pi+\frac{\pi}{3} \Rightarrow \mathrm{x}=\frac{\pi}{3}$
Put $n=1 \Rightarrow x=\pi+\frac{\pi}{3} \Rightarrow x=\frac{4 \pi}{3} \Rightarrow x=\frac{4 \pi}{3}$
So principal solution is $x=\frac{\pi}{3}$ and $\frac{4 \pi}{3}$
We know that $\tan \theta \times \cot \theta=1$
So $\cot x=\sqrt{3} \Rightarrow \tan x=\frac{1}{\sqrt{3}}$
The formula used: $\tan \theta=\tan \alpha \Rightarrow \theta=n \pi \pm \alpha, n \in I$
By using the above formula, we have
$\tan x=\frac{1}{\sqrt{3}}=\tan \frac{\pi}{6} \Rightarrow \theta=n \pi+\alpha, n \in 1$
Put $n=0 \Longrightarrow x=n \pi+\frac{\pi}{6} \Longrightarrow x=\frac{\pi}{6}$
Put $n=1 \Longrightarrow x=\pi+\frac{\pi}{6} \Rightarrow x=\frac{7 \pi}{6}$
So principal solution is $x=\frac{\pi}{6}$ and $\frac{7 \pi}{6}$
(v) Given: $\operatorname{cosec} x=2$
We know that $\operatorname{cosec} \theta \times \sin \theta=1$
So $\sin x=\frac{1}{2}$
Formula used: $\sin \theta=\sin \alpha \Rightarrow \theta=\mathrm{n} \pi+(-1)^{\mathrm{n}} \alpha, \mathrm{n} \in$
By using above formula, we have
$\sin x=\frac{1}{2}=\sin \frac{\pi}{6} \Rightarrow \theta=n \pi+\frac{\pi}{6}(-1)^{n}$
Put $n=0 \Longrightarrow \theta=0 \times \pi+\frac{\pi}{6}(-1)^{0} \Rightarrow \theta=\frac{\pi}{6}$
Put $\mathrm{n}=1 \Rightarrow \theta=1 \times \pi+\frac{\pi}{6}(-1)^{1} \Rightarrow \theta=1 \times \pi+\frac{\pi}{6}(-1)^{1} \Rightarrow \theta=\pi-\frac{\pi}{6}=\frac{5 \pi}{6}$
So principal solution is $x=\frac{\pi}{6}$ and $\frac{5 \pi}{6}$
(vi) Given: $\sec x=\frac{2}{\sqrt{3}}$
We know that $\sec \theta \times \cos \theta=1$
So $\cos x=\frac{\sqrt{3}}{2}$
Formula used: $\cos \theta=\cos \alpha \Rightarrow \theta=2 \mathrm{n} \pi \pm \alpha, \mathrm{n} \in \mid$
By using the above formula, we have
$\cos x=\frac{\sqrt{3}}{2}=\cos \frac{\pi}{6} \Rightarrow x=2 n \pi \pm \alpha, n \in 1$
Put $n=0 \Longrightarrow x=2 n \pi \pm \frac{\pi}{6} \Rightarrow x=\frac{\pi}{6}$
Put $n=1 \Rightarrow x=2 \pi \pm \frac{\pi}{6} \Rightarrow x=\frac{11 \pi}{6}, \frac{13 \pi}{6} \Rightarrow x=\frac{11 \pi}{6}, \frac{13 \pi}{6}$
$\left[\frac{13 \pi}{6}>2 \pi\right.$ So it is not include in principal solution $]$
So principal solution is $x=\frac{\pi}{6}$ and $\frac{11 \pi}{6}$
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All Study Material
- JEE Main
- Exam Pattern
- Previous Year Papers
- PYQ Chapterwise
- Physics
- Kinematics 1D
- Kinemetics 2D
- Friction
- Work, Power, Energy
- Centre of Mass and Collision
- Rotational Dynamics
- Gravitation
- Calorimetry
- Elasticity
- Thermal Expansion
- Heat Transfer
- Kinetic Theory of Gases
- Thermodynamics
- Simple Harmonic Motion
- Wave on String
- Sound waves
- Fluid Mechanics
- Electrostatics
- Current Electricity
- Capacitor
- Magnetism and Matter
- Electromagnetic Induction
- Atomic Structure
- Dual Nature of Matter
- Nuclear Physics
- Radioactivity
- Semiconductors
- Communication System
- Error in Measurement & instruments
- Alternating Current
- Electromagnetic Waves
- Wave Optics
- X-Rays
- All Subjects
- Physics
- Motion in a Plane
- Law of Motion
- Work, Energy and Power
- Systems of Particles and Rotational Motion
- Gravitation
- Mechanical Properties of Solids
- Mechanical Properties of Fluids
- Thermal Properties of matter
- Thermodynamics
- Kinetic Theory
- Oscillations
- Waves
- Electric Charge and Fields
- Electrostatic Potential and Capacitance
- Current Electricity
- Thermoelectric Effects of Electric Current
- Heating Effects of Electric Current
- Moving Charges and Magnetism
- Magnetism and Matter
- Electromagnetic Induction
- Alternating Current
- Electromagnetic Wave
- Ray Optics and Optical Instruments
- Wave Optics
- Dual Nature of Radiation and Matter
- Atoms
- Nuclei
- Semiconductor Electronics: Materials, Devices and Simple Circuits.
- Chemical Effects of Electric Current,