70 баллов! 1) sinx=a^2-1 2)0.5-sin(11p/12)

Question

70 баллов! 1) sinx=a^2-1 2)0.5-sin(11p/12)

Comments

what's the question?

---

sin(x)=a^2-1

---

2-1 это вся степень?

---

Нет, а во второй степени

---

no 1-sqrt2-sqrt3/2

Answer 1

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   * *

Answer 2

I hope this helps you

---

---

---

---

Comments

but i don't understand what you want

---

ask again but completely

---

tg(9p/8)-1==tg((p/8)+p)-1=-tg(p/8)-1<0answer: minusIt's answers other variant, but tasks are similar, it can help you

---

check now please

---

I need a solution of another example (0.5-sin(11p/12))the answer should be something like I wrote above in the comments (where I wrote a similar example and its solution)

---

ok i add it again

---

i hope, it's will be true

---

I'm sure my answers:)

---

are correct :)

---

thanks)