Given SiNx = 2cosx, find three angle function values of angle X

Given SiNx = 2cosx, find three angle function values of angle X

∵sinx=2cosx
∴tanx=2 cotx=1/2
And ∵ sinx2 + cosx2 = 1, replace SiNx = 2cosx
∴cosx=±√5/5 ∴sinx=±2√5/5
(cosx SiNx with positive or negative) formula is better to write step by step, which is more complete

Sorry... Given SiNx = 2cosx, find six trigonometric function values of angle X Thank you.

Square on both sides
(sinx)^2=4(cosx)^2
Add (cosx) ^ 2 to each side
1=5(cosx)^2
Cosx = root 5 / 5 or - root 5 / 5
sinx=2cosx
tanx= 1/2
cotanx = 2

Given SiNx = 2cosx, find the trigonometric function value of angle X

Because SiNx = 2cosx, (SiNx) ^ 2 + (cosx) ^ 2 = 1
So 4 (cosx) ^ 2 + (cosx) ^ 2 = 1
So cosx = plus or minus 5 / 5
So cosx = √ 5 / 5 SiNx = 2 √ 5 / 5 TaNx = 2
Or cosx = - √ 5 / 5 SiNx = - 2 √ 5 / 5 TaNx = 2

Ask the gods of y = 2sinx SiNx + 1 to help

Let t = SiNx (- 1 ≤ t ≤ 1), so y = 2t-t + 1. From the image of quadratic function: symmetry axis X = 1 / 4, opening up. Therefore, in the range [- 1,1], the minimum value of the function is f (1 / 4) = 7 / 8, and the maximum value is f (- 1) = 4

The range of y = (2sinx + 1) / (sinx-1)

Let SiNx = t
y=(2t+1)/(t-1)
=(2t-2+3)/(t-1)
=2+3/(t-1)
-1<=t<1
So - 2 < = T-1 < 0
3/(t-1)<=-3/2
The range of y = (2sinx + 1) / (sinx-1) (∞, 1 / 2]

f(x)=2sinx/2(sinx/2+cosx/2)-1 (1) Convert f (x) into sine function and write out the range of function (2) If α is an inner angle of a triangle, cut f (α + π / 4) = 1 to find α This kind of conversion is not likely

solution
f(x)=2sinx/2(sinx/2+cosx/2)-1
=2sin²x/2+2sinx/2cosx/2-1
=sinx-(1-2sin²x/2)
=sinx-cosx
=√2sin(x-π/4)
∵-1≤sin(x-π/4)≤1
∴-√2≤f(x)≤√2
The value range is [- √ 2, √ 2]
a∈(0,π)
f(a+π/4)=√2sin(a+π/4-π/4)=1
That is, Sina = √ 2 / 2
ν a = π / 4 or a = 3 π / 4

It is known that f (x) = 2sinx (cosx SiNx), where x belongs to R In the triangle ABC, the corresponding sides of a, B and C are a, B, C, f (a) = 0, B = 4, s triangle ABC = 6. Find the length of A

f(A)=2sinA(cosA-sinA)=0
We can get Sina = cosa
Angle a = 45 degrees
S triangle ABC = 1 / 2bcina = 6
C = 3 root sign 2
cosA=（b^2+c^2-a^2）/2bc
A = root 10

(x) = / (sinf) = / (x) = (x) = (x) = (x) = (x) = (x) = (x) = (x) = (x) = / (x) = (x) = (x) = / (x) = / (x) = (x)

1. 2. Y = (SiNx + 1) / (cosx + 2) ycosx + 2Y = SiNx + 1ycosx - SiNx = 1 - 2Y [√ (y 2 + 1)] sin (x + W) = 1 - 2ysin (...)

The value range of F (x) = 1 / 2 | SiNx + cosx | - 1 / 2 | SiNx cosx |, f (x)

[- 1, radical 2 / 2]
When sin > cos, f (x) = cosx
When cos > sin, f (x) = SiNx
SiNx > cosx and cosx > SiNx
X has a corresponding limit
Drawing, clear at a glance

Given the vector M = (cosx, - SiNx), n = (cosx, sinx-2 radical sign 3cosx), X belongs to R, Let f (x) = m * n

f(x)
=m.n
=(cosx,-sinx).(cosx,sinx-2√3cosx)
=(cox)^2- (sinx)^2 + 2√3sinxcosx
= cos2x + √3sin2x
= 2( (1/2)cos2x + √3/2 sin2x)
= 2( sin(2x+π/6)