Ex 3.4 Trigonomety ncert maths solution class 11

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 EXERCISE 3.4(Trigonometric function)

Find the principal and general solutions of the following equations(Ex 3.4 Trigonomety ncert maths solution class 11)

Question 1: \tan x=\sqrt{3}

Solution: Given, \tan x=\sqrt{3}

Since ‘\tan x‘ positive in Ist and IIIrd quadrant

Then, x = \frac{\pi}{3},\pi+\frac{\pi}{3}

\Rightarrow x = \frac{\pi}{3},\frac{4\pi}{3}

Hence the principal solution are \frac{\pi}{3},\frac{4\pi}{3}

Given, \tan x=\sqrt{3}

\tan x =\sqrt{3}

\Rightarrow \tan x= \tan \frac{\pi}{3}

Hence the general solution is

x = n\pi+\frac{\pi}{3} \text{ where } n \in Z

Question 2: \sec x = 2

Solution: Given, \sec x = 2

Since ‘\sec x‘ positive in Ist and IVth quadrant

Then, x = \frac{\pi}{3},2\pi-\frac{\pi}{3}

x = \frac{\pi}{3},\frac{5\pi}{3}

Since, \sec x =\frac{1}{\cos x}

Then, \cos x =\frac{1}{2}

\Rightarrow \cos x = \cos \frac{\pi}{3}

General solution:

x = 2n\pi\pm \frac{\pi}{3}\text{ where } n \in Z

Question 3: \cot x = -\sqrt{3}

Solution: Given, \cot x = -\sqrt{3}

Since, \cot x is negative in IInd and IVth quadrant

Then,x = \pi-\frac{\pi}{6},2\pi-\frac{\pi}{6}

\Rightarrow x = \frac{5\pi}{6},\frac{11\pi}{6}

Given, \cot x = -\sqrt{3}

\Rightarrow \tan x = -\frac{1}{\sqrt{3}}=\tan(\frac{5\pi}{6})

General solution:

x = n\pi+\frac{5\pi}{6}\text{ where } n \in Z

Question 4: \operatorname{cosec} x = -2

Solution: Given,\operatorname{cosec} x = -2

\operatorname{cosec} x is negative in IIIrd and IVth quadrant

Then, x = \pi+\frac{\pi}{6},2\pi-\frac{\pi}{6}

\Rightarrow x = \frac{7\pi}{6},\frac{11\pi}{6}

Again, given that \operatorname{cosec} x = -2

\Rightarrow \sin x = -\frac{1}{2}

\Rightarrow \sin x = \sin \frac{7\pi}{6}

Then general solution:

x = n\pi+(-1)^n\frac{7\pi}{6} \text{ where } n \in Z

Find the general solution for each of the following equations:

Question 5: \cos 4x = \cos 2x

Solution: Given \cos 4x = \cos 2x

General solution:

Using formula: \cos x = \cos y \Rightarrow x = 2n\pi\pm y

Hence, 4x = 2n\pi \pm 2x \text{ where } n \in Z

Taking ‘+’ve sign

4x = 2n\pi+2x

\Rightarrow 4x -2x= 2n\pi

\Rightarrow 2x = 2n\pi

\Rightarrow x = n\pi,\text{ where } n \in Z

Taking ‘-‘ve sign

4x = 2n\pi-2x

\Rightarrow 4x +2x= 2n\pi

\Rightarrow 6x = 2n\pi

\Rightarrow x = \frac{n\pi}{3},\text{ where } n \in Z

Question 6: \cos 3x + \cos x - \cos 2x = 0

Solution: Given, \cos 3x + \cos x - \cos 2x = 0

\Rightarrow 2\cos\frac{3x+x}{2}\cos \frac{3x-x}{2}-\cos 2x =0

\Rightarrow 2\cos2x\cos x-\cos 2x=0

\Rightarrow \cos 2x(2\cos x-1)=0

\Rightarrow \cos 2x = 0 \text{ or }2\cos x-1=0

\Rightarrow \cos 2x \text{ or } \cos x = \frac{1}{2}

Takaing \cos 2x =0

\Rightarrow \cos 2x = \cos \frac{\pi}{2}

\Rightarrow 2x = 2n\pi \pm \frac{\pi}{2}

\Rightarrow x = n\pi \pm \frac{\pi}{4} \text{ where } n \in Z

Taking, \cos x = \frac{1}{2}

\Rightarrow \cos x = \cos \frac{\pi}{3}

\Rightarrow x =2n\pi \pm \frac{\pi}{3}\text{ where } n \in Z

Question 7: \sin 2x + \cos x = 0

Solution: Given, \sin 2x + \cos x = 0

\Rightarrow 2\sin x\cos x+\cos x=0

\Rightarrow \cos x(2\sin x+1)=0

\Rightarrow \cos x = 0\text{ or } 2\sin x +1=0

\Rightarrow \cos x =\cos \frac{\pi}{2} \text{ or } \sin x = -\frac{1}{2}

\Rightarrow x = (2n+1)\frac{\pi}{2} \text{ or } \sin x = \sin\frac{7\pi}{6}

\Rightarrow x = (2n+1)\frac{\pi}{2} \text{ or } x = n\pi+(-1)^n \frac{7\pi}{6},\text{ where } n \in Z

Question 8: \sec^2 2x = 1- \tan 2x

Solution: Given, \sec^2 2x = 1- \tan 2x

\Rightarrow 1+\tan^22x=1-\tan 2x

\Rightarrow \tan^2x+\tan 2x=0

\Rightarrow \tan 2x(\tan 2x+1)=0

\Rightarrow \tan 2x =0 \text{ or } \tan 2x +1 =0

\Rightarrow \tan 2x = \tan 0\text{ or } \tan 2x = -1

\Rightarrow 2x = n\pi +0 \text{ or } \tan 2x= \tan \frac{3\pi}{4}

\Rightarrow x = \frac{n\pi}{2}\text{ or } 2x = n\pi+\frac{3\pi}{4}

\Rightarrow x = \frac{n\pi}{2}\text{ or } x = \frac{n\pi}{2}+\frac{3\pi}{8},\text{ where } n \in Z

Question 9: \sin x + \sin 3x + \sin 5x = 0

Solution: Given, \sin x + \sin 3x + \sin 5x = 0

\Rightarrow \sin 5x +\sin x+\sin 3x=0

\Rightarrow 2\sin\frac{5x+x}{2}\cos\frac{5x-x}{2}+\sin 3x=0

\Rightarrow 2\sin 3x\cos 2x +\sin 3x=0

\Rightarrow \sin 3x(2\cos 2x +1)=0

\Rightarrow \sin 3x = 0 \text{ or }2\cos 2x +1 =0

\Rightarrow \sin 3x = \sin 0 \text{ or } \cos 2x = -\frac{1}{2}

\Rightarrow 3x = n\pi +(-1)^n{0 }\text{ or } \cos 2x = \cos \frac{2\pi}{3}

\Rightarrow x =\frac{n\pi}{3}\text{ or } 2x = 2n\pi\pm\frac{2\pi}{3}

\Rightarrow x = \frac{n\pi}{3}\text{ or } x = n\pi\pm \frac{\pi}{3}\text{ where } n \in Z

 

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