If SiNx + cosx / SiNx cos = 2, then sin2x =? sinx+cosx/sinx-cos How to do this formula? Is there any formula

If SiNx + cosx / SiNx cos = 2, then sin2x =? sinx+cosx/sinx-cos How to do this formula? Is there any formula

solution
Deformation, can be obtained
sinx+cosx=2sinx-2cosx
sinx=3cosx
tanx=3
simple form
=（2sinxcosx)/(sin²x+cos²x)
=(2tanx)/(1+tan²x)
=6/10
=3/5

Given SiNx = 2cosx, find sin θ · cos θ

0

0

sinx=2cosx
tanx= 2
x is in 1st or 3rd quad.
sinx = ±2/√5
cosx =±1/√5
sinxcosx = 2/5
(sinx)^2 + 1/2 - sinxcosx
=2/5 + 1/2 - 2/5
=1/2

Find the square of y = 2cosx + the maximum value of sinx-4, and find the value of X

When SiNx - 1 / 4 = 0, y has a maximum value of - 15 / 8, where x = arcsin (1 / 4) + 2K π (k is an integer)

Y = 2cosx / 2 (SiNx / 2 + cosx / 2) simplification

y=2(cosx/2)(sin(x/2)+cos(x/2))
=sinx+2(cos(x/2))^2
=sinx+cosx+1
=√2sin(x+π/4)+1

Find the definition domain of function y = 2cosx / SiNx cosx SiNx cosx ≠ 0 i want to know how √ 2 (√ 2 / 2sinx - √ 2 / 2cosx) ≠ 0 and √ 2 (x - π / 4) ≠ 0 come from

The definition domain is {x | x ≠ K π + π / 4}, where k is an integer
Add:
sinx－cosx=√2[sinxcos(π/4)－cosxsin(π/4)]=√2sin(x－π/4).

Find the definition domain of function y = 2cosx / (SiNx cosx) SiNx cosx ≠ 0 √ 2Sin (x - π / 4) ≠ 0

sinx-cosx
=√2(√2/2*sinx-√2/2cosx)
=√2(sinxcosπ/4-cosxsinπ/4)
=√2sin(x-π/4)

Find the definition domain of y = 2cosx / SiNx cosx

0

0

It should be sin (30-x). If you do that, you take X as Y. The positive difference formula is SINXCOSY-COSXSINY=SIN (X-Y). Note that you do the original formula =SIN (Y-X).TGY=B/A= root sign 3/3.Y=30 degrees

Simplify 1 / 2cosx radical 3 / 2sinx

1 / 2cosx radical 3 / 2sinx
=cos60*cosx-sin60*sinx
=cos(x+60)