Question

Решите неравенство sinx=>cosx

Answer 1

Дано неравенство sinx=>cosx.
Исходим из равенства  cos²x = 1 - sin²x.
В исходном неравенстве возводим в квадрат обе части:
sin²x >= 1 - sin²x,
2sin²x >=1,
sin²x >= (1/2),
sinx >= +-√(1/2) >=+-1/√2 >= +-√2/2.
Отсюда х >= 2πn+(π/4),
              x <= 2πn-(3π/4).

График дан в приложении.

Answer 2

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To answer your question, the inequality sinx>=cosx can be solved by using the following steps:

- Rewrite the inequality as sinx-cosx>=0 - Use the identity sinx-cosx=\\sqrt{2}\\sin(x-\\frac{\\pi}{4}) - Solve the inequality \\sqrt{2}\\sin(x-\\frac{\\pi}{4})>=0 - Find the intervals where the sine function is non-negative - Add or subtract multiples of 2\\pi to find the general solution

The final answer is x\\in[\\frac{\\pi}{4}+2k\\pi,\\frac{5\\pi}{4}+2k\\pi), k\\in\\mathbb{Z}

You can also use online calculators to check your answer or get more details. For example, you can try [this one](https://ru.symbolab.com/solver/inequalities-calculator) or [this one](https://www.kontrolnaya-rabota.ru/s/neravenstva/). Alternatively, you can watch [this video](https://www.youtube.com/watch?v=VAxc_g8ZzLs) that explains how to solve a similar inequality. I hope this helps.