11 Июл 2019 в 19:43
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The first equation is a quadratic equation in terms of sine of x. To solve it, we can set sin x as a variable, say u, then we can rewrite the first equation as:

2u^2 - 5u + 2 = 0

This is a quadratic equation that can be factored as:

(2u - 1)(u - 2) = 0

Setting each factor to zero gives:

2u - 1 = 0 => u = 1/2
u - 2 = 0 => u = 2

Since sin x = u, the solutions for sin x are 1/2 and 2, but since the range of sine function is between -1 and 1, the only valid solution is sin x = 1/2.

Now, to find the corresponding values of x, we know that sin(30 degrees) = 1/2, so x = 30 degrees or x = pi/6 in radians.

The second equation involves sin x and y. You need more information to find the unique solutions.

20 Апр 2024 в 23:20
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