Сначала:
arcCos√3/2 = π/6
arcSin√2/2 = π/4
arc tg √3 = π/3
Теперь решаем:
а) Cos(π + π/6) = -Сosπ/6 = -√3/2
б) Cos(π/2 - π/3) =Sinπ/3 = √3/2
в) 8Sin x = 7Cos x |: Сosx ≠0
8tg x = 7
tgx = 7/8
x = arctg(7/8) + πk, k∈Z
<h3>
![\sqrt{x}+2\sqrt[4]{x}-8=0\\\\](https://tex.z-dn.net/?f=%5Csqrt%7Bx%7D%2B2%5Csqrt%5B4%5D%7Bx%7D-8%3D0%5C%5C%5C%5C)
</h3><h3>Пусть ⁴√x = a , a ≥ 0 , тогда</h3><h3>
![a^{2}+2a-8=0\\\\D=2^{2}-4*(-8)=4+32=36\\\\a_{1} =\frac{-2-6}{2}=\frac{-8}{2}=-4\\\\a_{2}=\frac{-2+6}{2}=\frac{4}{2}=2\\\\](https://tex.z-dn.net/?f=a%5E%7B2%7D%2B2a-8%3D0%5C%5C%5C%5CD%3D2%5E%7B2%7D-4%2A%28-8%29%3D4%2B32%3D36%5C%5C%5C%5Ca_%7B1%7D%20%3D%5Cfrac%7B-2-6%7D%7B2%7D%3D%5Cfrac%7B-8%7D%7B2%7D%3D-4%5C%5C%5C%5Ca_%7B2%7D%3D%5Cfrac%7B-2%2B6%7D%7B2%7D%3D%5Cfrac%7B4%7D%7B2%7D%3D2%5C%5C%5C%5C)
</h3><h3>
![a=2\\\\\sqrt[4]{x}=2\\\\x=2^{4}=16\\\\](https://tex.z-dn.net/?f=a%3D2%5C%5C%5C%5C%5Csqrt%5B4%5D%7Bx%7D%3D2%5C%5C%5C%5Cx%3D2%5E%7B4%7D%3D16%5C%5C%5C%5C)
</h3><h3><u><em>ОТВЕТ: 16</em></u></h3><h3><u><em /></u></h3>
3(2+x)>4-x
6+3x>4-x
3x+x>4-6
4x>-2
x>-1/2
x>-0,5
√(50a²)=-5a√2,a≤0
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3b² - 48 = 3( b² - 16 ) = 3( b - 4 )( b + 4 )
19x² - 19y² = 19( x - y )( x + y )
mx - mc² = m( x - c )( x + c )
- 4y² + 16 = - 4( y - 4 )( y + 4 )
81x^4 - 9x^2 = 9x²( 9x² - 1 ) = 9x²( 3x - 1 )( 3x + 1 )
a^6 - a^8 = a^6( 1 - a^2 ) = a^6( 1 - a )( 1 + a )