Solveeit Logo

Question

Question: If \(\cos 3x = - 1\) when \(0 \geqslant x \geqslant 360^\circ \)then \(x\) A. \(60^\circ ,180^\cir...

If cos3x=1\cos 3x = - 1 when 0x3600 \geqslant x \geqslant 360^\circ then xx
A. 60,180,30060^\circ ,180^\circ ,300^\circ
B. 180180^\circ
C. 60,18060^\circ ,180^\circ
D. 180,300180^\circ ,300^\circ

Explanation

Solution

We know that for cosθ=1\cos \theta = - 1 the general solution will be θ=(2n+1)π\theta = (2n + 1)\pi where n is the whole number and n0,1,2,3........n \in 0, 1, 2, 3........
Here we have θ=3x\theta = 3x and now by putting the value of nn we get the suitable xx.

Complete step by step solution:
Here we are given the equation that cos3x=1\cos 3x = - 1
And we know that cos180=1\cos 180^\circ = - 1
So we can write that cosπ=1\cos \pi = - 1
So cos3x=cosπ\cos 3x = \cos \pi
Now we can use the formula that ifcosθ=cosα\cos \theta = \cos \alpha , then θ=2nπ±α\theta = 2n\pi \pm \alpha
So here we are given that θ=3x,α=π\theta = 3x,\alpha = \pi
So we get that
3x=2nπ±π3x = 2n\pi \pm \pi
This is the general solution.
Now we get that x=(2n±1)π3x = (2n \pm 1)\dfrac{\pi }{3}
Here n is the whole number n0,1,2,3........n \in 0,1,2,3........
Here 2n±12n \pm 1 is the odd multiple of π\pi therefore we can also write that
x=(2n+1)π3x = (2n + 1)\dfrac{\pi }{3} and where n belongs to the integer
Which means that n=(0,±1,±2,±3,..........)n = (0, \pm 1, \pm 2, \pm 3,..........)
Now we are given that 0x3600 \geqslant x \geqslant 360^\circ
So we can say that x=(2n+1)π3x = (2n + 1)\dfrac{\pi }{3}
If n=0,x=π3,x=60n = 0,x = \dfrac{\pi }{3},x = 60^\circ
If n=1,x=π3,x=60n = - 1,x = - \dfrac{\pi }{3},x = - 60^\circ
So for n is negative we get x=60x = - 60^\circ therefore n cannot be negative
If n=1,x=π,x=180n = 1,x = \pi ,x = 180^\circ
If n=2,x=5π3,x=300n = 2,x = \dfrac{{5\pi }}{3},x = 300^\circ
If n=3,x=7π3,x=420n = 3,x = \dfrac{{7\pi }}{3},x = 420^\circ
As we can see that 420420^\circ is not in the range 0x3600 \geqslant x \geqslant 360^\circ

Therefore xx must be 60,180,30060^\circ ,180^\circ ,300^\circ for cos3x=1\cos 3x = - 1

Note:
If we are given that cosθ=cosα\cos \theta = \cos \alpha then its general solution will be given as θ=2nπ±α\theta = 2n\pi \pm \alpha
Similarly for sinθ=sinα\sin \theta = \sin \alpha its general solution is given as θ=nπ+(1)nα\theta = n\pi + {( - 1)^n}\alpha and for tanθ=tanα\tan \theta = \tan \alpha we have the general formula as θ=nπ+α\theta = n\pi + \alpha .