Square root as a function gives 2 values... x² = 1 has 2 values, 1 and -1 as they satisfy the equation x² - 1 = 0 x² - 1² = 0 (x-1)(x+1)=0 So, x=1 and x=-1 However, (root 1) = +1 only...
10 = 1010 (base 2) = 2^3 + 2^1 16^sin2x = 2^4sin2x 16^cos2x = 2^4cos2x 4sin2x = 3 or 1 Sin2x = 3/4 or 1/4 Sin x = 3^0.5/2 or 1/2 x = 60 deg or 30 deg In rad, x = pi/3 rad or pi/6 rad
те кто разбираются в математике вас поймут я простой человек и не заканчивал институт или университет но я в школе учился с математическом уклоном в г.Сумгаите.Вы абсолютно правы.Я очень хорошо разбирался в тригонометрии мне 54 если что и я до сих пор не занимаюсь математикой но не забыл
a beautiful question ! well now at 8'10 : sin²x = ¼. || sin²x = ¾. sin²x = (½)² || sin²x = (√¾)² a next step wouldn't it be at first : sin x = ±½ || sin x = ±√¾ x = ⅙π, ⅚π, 7π/6, 11π/6 || or x = ⅓π, ⅔π, 4π/3, 5π/3 , all of them of course ±2kπ . and only now put in the given restrictions 0 ≤ x ≤ ½π , so as allowed solutions will remain x = ⅙π, x = ⅓π . what we like to add on that moment is that we should not forget the two equations at first deliver four solutions each, so eight main values altogether . they all make their respective equations true, and after this the restraint just leaves two of them . perhaps this seems to be somehow exaggerated . well shouldn't mathematical solving ways principally lead to never overlooking any of the just possible solutions ?
@@robertcampomizzi7988 i did my bachelors electronics (Hons) in university 20 years ago and then masters in computers; but in India post 10th grade if you study maths you have to study trig and calculus so its radians for us.
16^sin²x+16^cos²x=10 since sin²x=(1−cos2x)/2 and cos²x=(1+cos2x)/2 4¹⁻ᶜᵒˢ²ˣ+4¹⁺ᶜᵒˢ²ˣ=10 4ᶜᵒˢ²ˣ+4⁻ᶜᵒˢ²ˣ=5/2 taking the logarithm of 4 eᶜᵒˢ²ˣ ˡᵒᵍ⁴+e⁻ᶜᵒˢ²ˣ ˡᵒᵍ⁴=5/2 (eᶜᵒˢ²ˣ ˡᵒᵍ⁴+e⁻ᶜᵒˢ²ˣ ˡᵒᵍ⁴)/2=5/4 since cos(ix)=(e⁻ˣ+eˣ)/2 cos i(log4·cos2x)=5/4 i(log4·cos2x)=−2nπ∓cos⁻¹(5/4) 2log2·cos2x=(2nπ±cos⁻¹(5/4))i (n∈ℤ) cos2x=(2nπ±cos⁻¹(5/4))i/2log2 2x=2mπ±cos⁻¹((2nπ±cos⁻¹(5/4))i/2log2) x=mπ±cos⁻¹((2nπ±cos⁻¹(5/4))i/2log2)/2 ({m, n}∈ℤ)
An interesting and challenging problem.
everybody: how detailed you are?
him: yes😂
When taking the square root, one must take into account both positive and negative values.
Square root as a function gives 2 values...
x² = 1 has 2 values, 1 and -1 as they satisfy the equation
x² - 1 = 0
x² - 1² = 0
(x-1)(x+1)=0
So, x=1 and x=-1
However, (root 1) = +1 only...
In the first quadrant sin and cos are positive.
Substitution is a superhero
Excellentissime 🙏🌹🇩🇿
for X= 0,π we have f(0) = 17>10 and for X= π/2 we have f(½) = 4+4 =8
10 = 1010 (base 2) = 2^3 + 2^1
16^sin2x = 2^4sin2x
16^cos2x = 2^4cos2x
4sin2x = 3 or 1
Sin2x = 3/4 or 1/4
Sin x = 3^0.5/2 or 1/2
x = 60 deg or 30 deg
In rad, x = pi/3 rad or pi/6 rad
👏👏👏
Ideal
Корень от квадрата числа нужно раскрывать с модулем. Кроме того, sin x - функция периодическая, а найдено только 2 решения от 0 до pi/2
те кто разбираются в математике вас поймут я простой человек и не заканчивал институт или университет но я в школе учился с математическом уклоном в г.Сумгаите.Вы абсолютно правы.Я очень хорошо разбирался в тригонометрии мне 54 если что и я до сих пор не занимаюсь математикой но не забыл
Ход решение обьсалютно правильно но ответы не правильные
В условии было задано, что Х от нуля до девяносто градусов.
thank you. Barquisimeto - VENEZUELAA
Excelente desarrolló.
Let 16^cos²x = u
( If 0 < x < π/2 then 0 < cosx < 1 )
So 16/u + u = 10
u² - 10u + 16 = 0
u = 8 or u = 2
16^cos²x = 8
4.cos²x = 3
cosx = √3/2
*x = π/6*
16^cos²x = 2
4.cos²x = 1
cosx = 1/2
*x = π/3*
Easy by hit and trial
Question has simple values so we can easily try the values 30degree and 60 degree are answers
But your method is awesome
nice
Спасибо очень понятно и подробно
Слмшком много действий. Сразу заменяем косинус через тождество и делаем замену, решаем уоавнение и рбратная замена. Автор сдишкомтмного пишет.
Good👍
Почему ответ в градусах? Должно быть число.
Atleast she explained in class 10th format
a beautiful question !
well now at 8'10 :
sin²x = ¼. || sin²x = ¾.
sin²x = (½)² || sin²x = (√¾)²
a next step wouldn't it be at first :
sin x = ±½ || sin x = ±√¾
x = ⅙π, ⅚π, 7π/6, 11π/6 ||
or
x = ⅓π, ⅔π, 4π/3, 5π/3 ,
all of them of course ±2kπ .
and only now put in the given restrictions
0 ≤ x ≤ ½π ,
so as allowed solutions will remain
x = ⅙π, x = ⅓π .
what we like to add on that moment is that we should not forget the two equations at first deliver four solutions each, so eight main values altogether . they all make their respective equations true,
and after this the restraint just leaves two of them .
perhaps this seems to be somehow exaggerated . well shouldn't mathematical solving ways principally lead to never overlooking any of the just possible solutions ?
you wrote it correctly
Absolutely correct ❤
Merci
How did you take 4y/4=1/4 ?
May be 'x'=sin (inverse )root 3/2, sin(inverse )1/root 2, cos(inverse )1/root2 and
😮 I like the sine way.
I prefer the method of your answering
x= kπ/6 where k is an integer and do not count 3, so k could be +/-1,2,4,5,7,8,....
Why did I get 60 and 45? I tried it using cos^2 (x) as basis
I tried using logarithm, but i got stuck when it became 1.2(sin²x+cos²x)=1 which is eventually 1.2=1 which is incorrect 😂
Should it be solved though log ??
Easy one😊 but here why negative values are taken?? I thing negative values have to be there
W2x +Wx+1=x
X=?
2x
X
This number is the foundation of omega
Мне рука нравится😅
He forgot the absolute value its x=[30, -30, 60, -60] * 2kpi k=natural constant
Where will I ever use this in non math, non STEM working life???
This has infinite solutions : npi +- pi/4
This went so well until u used degrees instead of radians
Let me guess.... you're studying engineering?
@@robertcampomizzi7988 i did my bachelors electronics (Hons) in university 20 years ago and then masters in computers; but in India post 10th grade if you study maths you have to study trig and calculus so its radians for us.
Cause most of the people are more comfortable in degree system
@@Dev_2267 How nonintellectual of them...thinking like civil or mechanical engineers and not physicists....
@@gauravverma5692 yes but reality is most of the people don't prefer mathematics stream people prefer biology over mathematics ☹️
Эти примеры нарочно что ли придумывают, для чего их надо решать?
α＋β=10
αβ=16
16^sin²x+16^cos²x=10
since sin²x=(1−cos2x)/2 and cos²x=(1+cos2x)/2
4¹⁻ᶜᵒˢ²ˣ+4¹⁺ᶜᵒˢ²ˣ=10
4ᶜᵒˢ²ˣ+4⁻ᶜᵒˢ²ˣ=5/2
taking the logarithm of 4
eᶜᵒˢ²ˣ ˡᵒᵍ⁴+e⁻ᶜᵒˢ²ˣ ˡᵒᵍ⁴=5/2
(eᶜᵒˢ²ˣ ˡᵒᵍ⁴+e⁻ᶜᵒˢ²ˣ ˡᵒᵍ⁴)/2=5/4
since cos(ix)=(e⁻ˣ+eˣ)/2
cos i(log4·cos2x)=5/4
i(log4·cos2x)=−2nπ∓cos⁻¹(5/4)
2log2·cos2x=(2nπ±cos⁻¹(5/4))i
(n∈ℤ)
cos2x=(2nπ±cos⁻¹(5/4))i/2log2
2x=2mπ±cos⁻¹((2nπ±cos⁻¹(5/4))i/2log2)
x=mπ±cos⁻¹((2nπ±cos⁻¹(5/4))i/2log2)/2
({m, n}∈ℤ)
計算間違ってたら指摘頼む
x=pi/4
√sin^2(x)=|sin(x)|
LHS = RHS ..30 degree
Khó quá
누가 저걸 60분법을 쓰냐 호도법 써야지
7줄이면 풀 수 있는 쉬운 문제를 너무 어렵게 풀었어.
I did it mentally in 20 seconds answer is 30 degrees or pi/6
No one cares.
Well, the full solution is
(⅙π)+2nπ and (⅓π)+2nπ (where n is the number of rotations)
Wow🤓🎉
@@kongshvalmagyar4984,
yeah quite smart .
and indeed no care that we might miss the
x = 60⁰ or π/3 . it's just for fun .