task/29655181
1. sinx = a² - 1 Уравнение имеет решение , если
- 1 ≤ a² - 1 ≤ 1 ⇔ 0 ≤ a² ≤ 2 ⇔ - √2 ≤ a ≤ √2 * * * a ∈ [ - √2; √2 ]
x =( -1)⁻ⁿ arcsin(a² - 1 ) + πn , n ∈ ℤ .
* * * a ∈ ( - ∞ ; - √2) ∪ (√2 ; ∞) → нет решения x ∈ ∅ * * *
2. 0,5 - sin(11π/12) =0,5 - sin(π -π/12) = 0,5 - sin(π/12) = 0,5*( 1 - √(2 -√3) ) , т.к.
sin(π/12) = √[ ( 1 -cos(π/6) ) /2] = 0,5√(2 -√3)
* * * √(2 -√3) =√( 2(4 - 2√3) / 4) = √( 2(√3 - 1)² /4 ) = 0,5 (√3 - 1)√2 * *
I hope this helps you
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task/29655181
1. sinx = a² - 1 Уравнение имеет решение , если
- 1 ≤ a² - 1 ≤ 1 ⇔ 0 ≤ a² ≤ 2 ⇔ - √2 ≤ a ≤ √2 * * * a ∈ [ - √2; √2 ]
x =( -1)⁻ⁿ arcsin(a² - 1 ) + πn , n ∈ ℤ .
* * * a ∈ ( - ∞ ; - √2) ∪ (√2 ; ∞) → нет решения x ∈ ∅ * * *
2. 0,5 - sin(11π/12) =0,5 - sin(π -π/12) = 0,5 - sin(π/12) = 0,5*( 1 - √(2 -√3) ) , т.к.
sin(π/12) = √[ ( 1 -cos(π/6) ) /2] = 0,5√(2 -√3)
* * * √(2 -√3) =√( 2(4 - 2√3) / 4) = √( 2(√3 - 1)² /4 ) = 0,5 (√3 - 1)√2 * *
=tg((p/8)+p)-1=-tg(p/8)-1<0
answer: minus
It's answers other variant, but tasks are similar, it can help you
the answer should be something like I wrote above in the comments (where I wrote a similar example and its solution)
Verified answer
I hope this helps you