log2 (6+x) < 3
5^((x^2) - 15) = 25^x
sinx=2 cosx
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Ответил Аноним
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1)log₂(6+x)<3
log₂(6+x)<log₂8
0<6+x<8
-6<x<2
x∈(-6;2)
![2)5^{ x^{2} -15}=25^{x} \
5^{ x^{2} -15}=5^{2x} \
x^{2} -15=2x \
x^{2} -2x-15=0 \
D=64;x_{1}=5;x_{2}=-3 \
3)sinx=2cosx (:cosx neq 0) \
tgx=2 \
x=arctg2+ pi k](https://tex.z-dn.net/?f=2%295%5E%7B+x%5E%7B2%7D+-15%7D%3D25%5E%7Bx%7D+%5C+%0A5%5E%7B+x%5E%7B2%7D+-15%7D%3D5%5E%7B2x%7D+%5C+%0A+x%5E%7B2%7D+-15%3D2x+%5C+%0A+x%5E%7B2%7D+-2x-15%3D0+%5C+%0AD%3D64%3Bx_%7B1%7D%3D5%3Bx_%7B2%7D%3D-3+%5C+%0A3%29sinx%3D2cosx+%28%3Acosx+neq+0%29+%5C+%0Atgx%3D2+%5C+%0Ax%3Darctg2%2B+pi+k%0A "2)5^{ x^{2} -15}=25^{x} \
5^{ x^{2} -15}=5^{2x} \
x^{2} -15=2x \
x^{2} -2x-15=0 \
D=64;x_{1}=5;x_{2}=-3 \
3)sinx=2cosx (:cosx neq 0) \
tgx=2 \
x=arctg2+ pi k")
k∈Z
log₂(6+x)<log₂8
0<6+x<8
-6<x<2
x∈(-6;2)
k∈Z
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