Question

3cos-1 x = cos-1 (4x3 - 3x), x∈[ 1/2, 1 ]

Answer 1

Answer:

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Answer 2

Answer:

Step-by-step explanation:

माना  $cos^{-1}x=\theta$  तब  

$x=cos \theta$और   0 ≤ Θ ≤ π

∵ x ∈ [ 1/2, 1 ],  इसलिए   $\dfrac{1}{2} \leq x\leq 1$

$=>cos\dfrac{\pi}{3} \leq cos\theta \leq cos0\\\\\\=> \dfrac{\pi}{3} \geq \theta \geq 0$

( ∵ cos Θ का मान  [0,π]  में घटता है। )

⇒ π ≥ 3Θ ≥ 0, i.e.. 3Θ ∈ [0,π]

$cos^{-1}(4x^3-3x)\\\\=cos^{-1}(4cos^3\theta-3cos\theta)\\\\=cos^{-1}(cos3\theta)\\\\=3\theta\\\\=3 cos^{-1}x$