Solve the equation 1 + SiNx + cosx + sin2x + cos2x = 0 It is necessary to specify the extraction process

Solve the equation 1 + SiNx + cosx + sin2x + cos2x = 0 It is necessary to specify the extraction process

1+sinx+cosx+sin2x+cos2x=0(1+sin2x)+(sinx+cosx)+(cos²x-sin²x)=0(sinx+cosx) ²+(sinx+cosx)+ (cosx+sinx) (cosx-sinx)=0(sinx+cosx)[(sinx+cosx)+1+(cosx-sinx)]=0(sinx+cosx)(2cosx+1)=0sinx+cosx...

If f (SiNx) = 3-cos2x, then f (cosx)=______ .

f（cosx）=f[sin(π
2−x)]
=3-cos（π-2x）
=3+cos2x．
So the answer is: 3 + cos2x

Is the period of the function f (x) = sin2x.cos2x? Odd or even

f(x)=sin2xcos2x=1/2*sin4x
So t = 2 π / 4 = π / 2
f(-x)=-1/2*sin4x=-f(x)
So it's an odd function

Is the function f (x) = sin2x * cos2x odd or even? What is the period

f(x)=1/2*(2sin2xcos2x)
=1/2*sin4x
So obviously f (- x) = - f (x)
The definition field is r, symmetric about the origin
So it's an odd function
T=2π/4=π/2

If f (x) = (1 + cos2x) sin ^ 3x, X ∈ R, then f (x) is () If f (x) = (1 + cos2x) sin ^ 3x, X ∈ R, then f (x) is () A. Odd functions with minimum positive period π B. Even functions with minimum positive period π C. Odd functions with minimum positive period π / 2 D. Even functions with minimum positive period π / 2 The cube of SiNx is SiNx

First, f (x) = (1 + cos2x) sin? X
f(-x)=[1+cos2（﹣x）] sin³（﹣x）
=(1+cos2x)·（﹣sin³x）
=﹣f(x)
/ / F (x) is an odd function! Exclude B and D options
Then, f (x) = (1 + cos2x) sin? X
=(1 + cos2x) · SiNx · (1-cos2x) / 2
=[（1+cos2x）·（1-cos2x）]／2 ·sinx
=sin²2x·sinx／2
=sin³2x／4cosx
The minimum positive period of the fractional molecule is π, and the minimum positive period of the denominator is π / 2, so the small π / 2 is taken
Select c

Is the function y = sin (1 / 3x + π / 2) odd or even?

y=sin(1/3x+π/2)=cosx/3.
Because cos (- X / 3) = cosx / 3
Therefore, y = sin (1 / 3x + π / 2) = cosx / 3 is even function

Is f (x) = sin (cos2x) even? Help me with another problem

Because f (- x) = sin (COS (- 2x)) = sin (cos2x) = f (x)
So it's even

The following functions are even functions: A, y = cos2x B, y = x ^ 3 + 1 C, y = x? 2 = x D, y = SiNx

A is right, f (x) = f (- x)

Y = SiNx cosx + 1 odd function or even function,

y=sinx - cosx +1
=Radical 2 * sin (x - π / 4) + 1
F (- x) = radical 2 * sin (- X - π / 4) + 1 is not equal to f (x)
It is neither odd nor even
If you can't, just bring a few numbers or draw with a Geometer's Sketchpad

Is y = SiNx cosx + 1 odd or even

∵f（-x）=sin(-x)-cos(-x)+1
=-sinx-cosx+1 ≠f（x）≠-f（x）
So this function is not even and odd
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