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Sagot :
[tex]\\ \rm\Rrightarrow cosx=cos2x[/tex]
[tex]\\ \rm\Rrightarrow cosx-cos2x=0[/tex]
[tex]\\ \rm\Rrightarrow cosx-2cos^2x+1=0
- Multiply -
[tex]\\ \rm\Rrightarrow 2cos^2x-cosx-1=0[/tex]
[tex]\\ \rm\Rrightarrow (cosx-1)(2cosx+1)=0[/tex]
[tex]\\ \rm\Rrightarrow cosx=1\:or\dfrac{-1}{2}[/tex]
[tex]\\ \rm\Rrightarrow x=cos^{-1}(1)\:or\:cos^{-1}(-1/2)[/tex]
[tex]\\ \rm\Rrightarrow x=\dfrac{2\pi}{3},0,\dfrac{4\pi}{3}[/tex]
Value of x is equals to (2π / 3) , ( 4π / 3) , 2π as 0 < x ≤ 2π.
What are trigonometric functions?
" Trigonometric functions are defined as the mathematical expression of representing the relation between the sides and angles of a right angled triangle."
Formula used
cos2θ = cos²θ - sin²θ
= 2cos²θ - 1
According to the question,
Given ,
0 < x ≤ 2π
cos x = cos 2x
Substitute the value of trigonometric function cos2x from formula,
cos x = 2cos²x - 1
⇒2cos²x - cos x - 1 = 0
Split the middle term of trigonometric equation we get,
2cos²x - 2cos x + cos x - 1 = 0
⇒2cos x ( cos x - 1) + 1( cos x - 1) = 0
⇒(cos x - 1)(2cos x + 1) = 0
⇒ cos x - 1 =0 or 2cos x + 1 =0
⇒cos x = 1 or cos x = (- 1 / 2)
cos x = 1 ⇒ x = 0 or 2π
But 0 < x ≤ 2π
Therefore,
cos x = 1 ⇒ x = 2π
[tex]cos (\pi -\frac{\pi }{3 }) = cos (\pi + \frac{\pi }{3 }) = -cos (\frac{\pi }{3 })=-\frac{1}{2}[/tex]
⇒ [tex]x = \frac{2\pi }{3 } and x = \frac{4\pi }{3 }[/tex]
Hence, Option(A) is the correct answer.
Learn more about trigonometric function here
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