f x 1 cos2x 8sin 2 x sin2x find the minimum value, please everyone, thank you

x should have a range, right? y=f(x)=(1+cos2x+8sin 2x) sin2x =(1+cos2x+4-4cos2x) sin2x =(5-3cos2x) sin2x =(5-3cos2x) [1-(cos2x) 2] and then solve according to the range Follow-up: When 0 is basically the same as the following.

cos2x=1-2sin 2x 2sin 2x=1-cos2x 8sin 2x=4-4cos2x y=f(x)=(1+cos2x+8sin 2x) sin2x =(1+cos2x+4-4cos2x) sin2x =(5-3cos2x) sin2x =(5-3cos2x) [1-(cos2x) 2] 00, sin2x>0 y>0 y* [1-(cos2x) 2]=5-3cos2x y 2*[1-(cos2x) 2]=(5-3cos2x) 2 (9+y 2)*(cos2x) 2-30(cos2x)+25-y 2=0 The unknown of the equation is (cos2x) 0, i.e., (-30) 2-4*(9+y 2)*(25-y 2) 0 y 4-16y 2 0 y 2*(y+4)*(y-4)) 0 y 4 (the other solution y -4 is rounded) Y min=4 y=4 (9+y 2)*(cos2x) 2-30(cos2x)+25-y 2=0 (5cos2x-3) 2=0 cos2x=3 5,sin2x=4 5 y=f(x)=(1+cos2x+8sin 2x) sin2x =(1+cos2x+4-4cos2x) sin2x =(5-3cos2x) sin2x =(5-3*3 5) (4 5) =4 Answer: When 0When

Summary. Square ah.

f(x)=1 sin2x++3 cos+2x.

Is there a result, that 2 is squared.

Square ah. Be.

This one is not. This one.

Good. Here are the results.

Take a look. Okay, thank you, teacher.

Teacher, are you busy, if you are not busy, can you help me organize this answer, if you are busy, forget it, hard teacher.

Good. Just rewrite it, right?

f(x)=2sin(x—α)cos(x-α)f(x)=sin[2(x—α)

If: when the nuclear field x=0, the circle collapse f(x)=sin[2(x— )0 then 2(x— )k

kπ/2

f(x)=(1+2cos²a-1+8sin²x)/(2sinxcosx)

2(cos²x+4sin²x)/(2sinxcosx)=cosx/sinx+4suinx/cosx=cotx+4tanx

x is an acute angle. So tanx>>0

f(x)2 (cotx*4tanx)=4, so the minimum value = 4

It's still mean!! Hahahaha, I'll finally do it!

The first is formed into a quadratic (that is, the exponent of the numerator and denominator is replaced with the same) (sin x+cos x+cos x+cos x-sin x+8sin x) (2sinxcosx).

Simplification obtains (4sin x+cos x) sinxcosx(4sinx) cosx+cosx sinx are all worth to f(x)>=4

s, c, and "2" represent sin, cos, and root number 2! 1), g(x)=[(s2x) 2]-(1 4), move up 1 4, get y=(s2x) to the left translation of 4, y= 2= 2=c(2x) 2=[(cx) 2-(sx) 2] 2=f(x). 3)、h(x)=[c2x/2)-[s2x/2)+(1/4)=(1/4)-(1/2)(s2x-c2x)。

Multiplication root number 2,=(1 4)-(2" 2)[(1 "2")s2x-(1 "2")c2x)]=(1 4)-(2" 2)[s2x·c(vultures 4)-c2x·s(vultures 4)]=(1 4)-(2" 2)s[2x-(vultures 4)]. 4) When s[2x-(vulture 4)]=1, the minimum value h(x)=(1 4)-(2" 2)=(1-2"2") 2.

In the first step, f(x) is simplified to obtain f(x)=2cos 2x sinxcosx-sin 2x).

In the second step, we know that cosx is not 0 from the definition domain, and f(x) is divided by the square of cosx, so that f(x)=2 tanx-tan 2x).

In the third step, g(x)=tanx-tan 2x, the range of tanx is (0,1 4) from the definition domain, and f(x) has a minimum value when g(x) obtains the maximum value.

b/2a=1/2 g(1/2)=1/4

The minimum value of f(x) is 8

f(x)=(cos2x+1)/(sinxcosx-sin^2x)=(cosx）^2/(sinxcosx-(sinx)^2);

Divide the same up and down by (cosx) 2 (which is not 0 from the inscription), and get f(x)=2 [tgx(1-tgx)], which is known by the inequality ab<=[a+b) 2] 2 (a, b>0).

When a=b, ab takes the maximum value and 1 ab takes the minimum value, so when tgx=1-tgx, f(x) takes the minimum value 8

Use the power formula: cos2x=(cosx) 2-(sinx) 2=2(cosx) 2-1 f(x)=[2*(cos 2)+8*(sin 2)] 2sinx*cosx)=(1+4tan 2) tan =(1 tanx)+4*tanx>=2*root number[(1 tanx)*4tanx]=4 So the minimum value is 4 Follow-up: Trouble to ask for 2*root number [(1 tanx)*4tanx] How did you launch it?

f(x)=cosx-1+cos x-2cos x+1+7 4 We assume t=cosx[-1,1].

f(t)=t-t²+7/4=-(t²-t+1/4-1/4)+7/4=-(t-1/2)²+2

So when t = 1 2, f max = 2

I don't know if I understand o( o ha!

This question is mainly tested for the exchange method, and if you don't understand, you can also ask (o).

f(x)=cosx-(1-cos²x)-(2cos²x-1)+7/4=cosx-1+cos²x-2cos²x+1+7/4=-(cosx-1/2)²+2

Because cosx is less than or equal to 1 and greater than or equal to -1, there is a maximum value when it is equal to 1 and 2, and because it cannot be combined here, cosx can only be regarded as a t and solved by the method of finding the maximum value of a quadratic function.