Question
Question: If \(\cos 3x = - 1\) when \(0 \geqslant x \geqslant 360^\circ \)then \(x\) A. \(60^\circ ,180^\cir...
If cos3x=−1 when 0⩾x⩾360∘then x
A. 60∘,180∘,300∘
B. 180∘
C. 60∘,180∘
D. 180∘,300∘
Solution
We know that for cosθ=−1 the general solution will be θ=(2n+1)π where n is the whole number and n∈0,1,2,3........
Here we have θ=3x and now by putting the value of n we get the suitable x.
Complete step by step solution:
Here we are given the equation that cos3x=−1
And we know that cos180∘=−1
So we can write that cosπ=−1
So cos3x=cosπ
Now we can use the formula that ifcosθ=cosα, then θ=2nπ±α
So here we are given that θ=3x,α=π
So we get that
3x=2nπ±π
This is the general solution.
Now we get that x=(2n±1)3π
Here n is the whole number n∈0,1,2,3........
Here 2n±1 is the odd multiple of π therefore we can also write that
x=(2n+1)3π and where n belongs to the integer
Which means that n=(0,±1,±2,±3,..........)
Now we are given that 0⩾x⩾360∘
So we can say that x=(2n+1)3π
If n=0,x=3π,x=60∘
If n=−1,x=−3π,x=−60∘
So for n is negative we get x=−60∘ therefore n cannot be negative
If n=1,x=π,x=180∘
If n=2,x=35π,x=300∘
If n=3,x=37π,x=420∘
As we can see that 420∘ is not in the range 0⩾x⩾360∘
Therefore x must be 60∘,180∘,300∘ for cos3x=−1
Note:
If we are given that cosθ=cosα then its general solution will be given as θ=2nπ±α
Similarly for sinθ=sinα its general solution is given as θ=nπ+(−1)nα and for tanθ=tanα we have the general formula as θ=nπ+α.