Question

1)2 cos^2x+sinx-2=0 2)3sin^7x-sinx-4=0

Answer 1

1)

$2cos^{2}x+sinx-2=0\\ 2(1-sin^{2}x) +sinx-2=0\\2-2sin^{2}x+sinx-2=0\\sinx(1-2sinx)=0$

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$x=\pi n$, $n$∈$Z$                  $sinx=\frac{1}{2}$

                                     $x=(-1)^{k}*\frac{\pi }{6} +\pi k$, $k$∈$Z$

2) тут точно ошибки нет?

Answer 2

To solve the given equations, let's take them one by one:

1. 2 cos^2(x) + sin(x) - 2 = 0

To solve this equation, let's first recognize that cos^2(x) can be replaced with (1 - sin^2(x)), using the identity: cos^2(x) + sin^2(x) = 1.

So, the equation becomes:

2(1 - sin^2(x)) + sin(x) - 2 = 0

Now, let's rearrange and combine like terms:

2 - 2sin^2(x) + sin(x) - 2 = 0

-2sin^2(x) + sin(x) = 0

Factor out sin(x):

sin(x) (-2sin(x) + 1) = 0

Now, set each factor to zero and solve for sin(x):

sin(x) = 0 x = arcsin(0) = 0 (plus multiples of 2π)

-2sin(x) + 1 = 0 -2sin(x) = -1 sin(x) = 1/2 x = arcsin(1/2) = π/6 (plus multiples of 2π) and x = 5π/6 (plus multiples of 2π)

So, the solutions to the equation are: x = 0, π/6, 5π/6 (plus multiples of 2π)

2. 3sin^7(x) - sin(x) - 4 = 0

This equation seems more complicated, but we can still attempt to solve it. Unfortunately, there is no simple closed-form solution for this type of equation. You could use numerical methods, such as Newton's method or graphical methods, to approximate the solutions. However, the exact solutions would involve complex expressions.

If you're working on a particular problem or task, there might be other ways to approach it that don't require finding exact solutions for this equation. If you need approximate values, numerical methods can be used with the help of software or calculators.