In △ ABC, C = B (1 + 2cosa), the proof angle a = 2 angle B Step by step, thank you

In △ ABC, C = B (1 + 2cosa), the proof angle a = 2 angle B Step by step, thank you

Because a / Sina = B / SINB = C / sinc = 2R
So we get it from the original formula
2RsinC = 2RsinB(1+2CosA)
sinC = sinB(1+2CosA)
also
sinC = sin (180 - A - B)=sin(A+B)
So sin (a + b) = SINB (1 + 2cosa)
sinAcosB + cosAsinB = sinB + 2sinBcosA
sinAcosB - cosAsinB = sinB
sin(A-B)=sinB
And 0 < A, B < 180
So A-B = B, then a = 2B

It is known that ABC is the equation a (1-x) + 2bx + C (1 + x) = 0 for the opposite side of angle a, B and C in triangle ABC
There are two equal real roots and 3C = a + 3B. ① judge the shape of triangle ABC. ② find the value of sina + SINB

1) A (1-x Square) + 2Ab + C (1 + x square) = 0 (C-A) x ^ 2 + 2Ab + A + C = 0 because there are two equal real roots, so 0-4 (C-A) (2Ab + A + C) = 0, so C = a, so it is isosceles triangle. 2) because 3C = a + 3b, so a = C = 3 / 2B, so Sina = 2 (radical 2) / 3, SINB = 4 (radical 2) / 9, so si

Given that the equation x ^ 2 (Sina) + 2xsinb + sinc = 0 has multiple roots, then the trilateral a, B and C of △ ABC satisfy the relation
A.b=ac B.b^2=ac C.a=b=c D.c=ab

B
It can be seen that the equation has multiple roots that the equation is a quadratic equation of one variable and the discriminant of the root is 0
That is, (2 × SINB) ^ 2-4 × Sina × sinc = 0
You can get it
(sinB)^2=sinA×sinC
So SINB / Sina = sinc / SINB
Combined with sine theorem, we can get
b/a=c/b
So B ^ 2 = AC

Let ABC be the internal angle of a triangle and satisfy the equation (SINB Sina) x ^ 2 + (Sina sinc) x + (sinc SINB) = 0, which has two equal real roots. Find the range of angle B

From the meaning of the title
(sinA-sinc)^2-4*(sinB-sinA)*(sinC-sinB)=0 (1)
It's too much trouble to expand this formula
From sine theorem
A = 2 * rsina
sinA=a/(2*R)
sinB=b/(2*R)
sinC=c/(2*R)
Take the above three formulas into (1)
have to
(a-c)^2-4*(b-a)*(c-b)=0
After simplification
a^2+2*a*c+c^2+4*b^2-4*a*b-4*b*c=0
(a+c)^2-4*b(a+c)+4*b^2=0
(a+c-2*b)^2=0
a+c=2*b
cosB=(a^2+c^2-b^2)/(2*a*c)=(a^2+c^2-((a+c)/2)^2)/(2*a*c)=3*(a^2+c^2)/(8*a*c)-1/4
>=(3*2*a*c)/(8*a*c)-1/4
That is, CoSb > = 1 / 2
So the range of B is (0, π / 3]