4-cos-2-x-4-cos-2-3x-cos-x-cos-2-3x-0-0-pi-2-Find-x-

Question Number 208945 by hardmath last updated on 28/Jun/24

4 \cos^{2} x - 4 \cos^{2} 3 x \cos x + \cos^{2} 3 x = 0

[0; \frac{π}{2}]

Find : x = ?

Answered by Berbere last updated on 28/Jun/24

4 {c o s}^{2} (x) + {c o s}^{2} (3 x) ⩾ 4 ∣ c o s (3 x) c o s (x) ∣⩾ 4 ∣ {c o s}^{2} (3 x) c o s (x) ∣

4 {c o s}^{2} (x) + {c o s}^{2} (3 x) - 4 {c o s}^{2} (3 x) c o s (x) = 0

\Rightarrow 4 ∣ c o s (3 x) c o s (x) ∣ - 4 {c o s}^{2} (3 x) c o s (x) = 0

\Rightarrow∣ c o s (3 x) c o s (x) ∣= {c o s}^{2} (3 x) c o s (x)

\Rightarrow∣ c o s (3 x) ∣ c o s (x) - {c o s}^{2} (3 x) c o s (x) = 0

c o s (x) (∣ c o s (3 x) ∣ - {c o s}^{2} (3 x)) = 0

x = \frac{π}{2}, o r ∣ c o s (3 x) ∣= c o s (3 x) . c o s (3 x)

\Rightarrow c o s (3 x) \in {- 1, 1}

c o s (3 x) \Rightarrow 3 x = k π

x = \frac{k π}{3}; \Rightarrow x = 0, \frac{π}{3}

a f t e r t c h e k] x \in {\frac{π}{3}, \frac{π}{2}}

Commented by hardmath last updated on 28/Jun/24

thank you dear professor

Commented by Berbere last updated on 28/Jun/24

W i t h e P l e a s u r G o d b l e s s Y o u

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