1.cos(x/2+n/4)+1=02.sin^2x-2cosx+2=03.sinx cosx+2sin^x-cos^2x=04.3sin^2x-4sin sosx+5cos^2x=25.sin3x=cos3x {0;4}

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For the first equation: cos(x/2 + n/4) + 1 = 0.Subtract 1 from both sides: cos(x/2 + n/4) = -1.

Since the cosine function ranges from -1 to 1, the only solution is when the angle inside the cosine function is for n = 2πk - 1, where k ∈ Ζ.

Therefore, x/2 + n/4 = 2πk - 1.
x/2 = 2πk - 1 - n/4
x = 4πk - 2 - n/2

For the second equation: 2sin^2x - 2cosx + 2 = 0.

Divide everything by 2:
sin^2x - cosx + 1 = 0.

In terms of cosine, sin^2x = 1 - cos^2x.
Substitute:
1 - cos^2x - cosx + 1 = 0
-cos^2x - cosx + 2 = 0.

This is a quadratic equation in terms of cos(x), which you can solve using the quadratic formula.

For the third equation: sinx cosx + 2sin^x - cos^2x = 0.

This can be simplified further by substituting sin^2x = 1 - cos^2x:

cosx(1 - cos^2x) + 2(1 - cos^2x) - cos^2x = 0
cosx - cos^3x + 2 - 2cos^2x - cos^2x = 0

This is a cubic equation in terms of cos(x), which can be solved using numerical methods.

For the fourth equation: 3sin^2x - 4sinx cosx + 5cos^2x = 2.

This is similar to the third equation but with additional terms.
You can rewrite this equation in terms of sin2x and cos2x and solve for x using trigonometric identities.

For the fifth equation: 5sin3x = cos3x.

Divide by cos3x and simplify:
5tan3x = 1
tan3x = 1/5.

Solving for x in the domain [0, 4], you can find the values of x that satisfy this equation.