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malivina76
@malivina76
October 2021
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Решите тригонометрическое неравенство:
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4(1-sin²x)-(2√2-2)-4+√2>0
4-4sin²x-(2√2-2)sinx-4+√2>0
4sin²x+(2√2-√)sinx-√2<0
sinx=a
4a²+(2√2-2)a-√2<0
D=8-8√2+4+16√2=8*8√2+4=(2√2+2)²
√D=2√2+2
a1=(2-2√2-2√2-2)/8=-√2/2
a2=(2-2√2+2√2+2)/8=1/2
-√2/2<a<1/2⇒-√2/2<sinx<1/2
x∈(-π/4+2πn;π/6+2πn) U (5π/6+2πn;5π/4+2πn)
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Answers & Comments
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4(1-sin²x)-(2√2-2)-4+√2>04-4sin²x-(2√2-2)sinx-4+√2>0
4sin²x+(2√2-√)sinx-√2<0
sinx=a
4a²+(2√2-2)a-√2<0
D=8-8√2+4+16√2=8*8√2+4=(2√2+2)²
√D=2√2+2
a1=(2-2√2-2√2-2)/8=-√2/2
a2=(2-2√2+2√2+2)/8=1/2
-√2/2<a<1/2⇒-√2/2<sinx<1/2
x∈(-π/4+2πn;π/6+2πn) U (5π/6+2πn;5π/4+2πn)