Ответ:
cos²x-sin²x-cos²x-√2sinx=0
-sinx*(sinx+√2)=0; sinx=-√2,∅, т.к. -1 ≤sinx≤1
sinx=0
x=πn; n∈Z
-π≤πn≤π
-1≤n≤1
n=-1; x=-π
n=0; х=0
n=1; х=π
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Ответ:
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cos²x-sin²x-cos²x-√2sinx=0
-sinx*(sinx+√2)=0; sinx=-√2,∅, т.к. -1 ≤sinx≤1
sinx=0
x=πn; n∈Z
-π≤πn≤π
-1≤n≤1
n=-1; x=-π
n=0; х=0
n=1; х=π