In a ΔABC , 3sinx -4 sin² x - k=0 , 0 Find the measure of
abhi178:
sir i think something missing
3sinx-4sin^3x-k=0
but here 3sinx-4sin^2x-k=0
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A , B , C are angle of triangle also this angle satisfied given equation.
now,
3sinA-4sin^3A-k=0 => sin3A=k
3sinB-4sin^3B-k=0 =>sin3B=k
3sinC-4sin^3C-k=0 => sin3C=k
here we see all
sin3A=sin3B=sin3C =k
sin3A=sin3B
sin3A=sin(180°-3B)
3A+3B=180°
A+B=60°----------(1)
but we know ,
sum of all angles of triangle is are 180°
e.g. A+B+C=180°
form equation (1)
60°+C=180°
C=120°
hence k=sin3C=sin360°=0
k=0
now,
3sinA-4sin^3A-k=0 => sin3A=k
3sinB-4sin^3B-k=0 =>sin3B=k
3sinC-4sin^3C-k=0 => sin3C=k
here we see all
sin3A=sin3B=sin3C =k
sin3A=sin3B
sin3A=sin(180°-3B)
3A+3B=180°
A+B=60°----------(1)
but we know ,
sum of all angles of triangle is are 180°
e.g. A+B+C=180°
form equation (1)
60°+C=180°
C=120°
hence k=sin3C=sin360°=0
k=0
abhi , here we don't have cube on second term
actually here i think mistake because traingle have three angle which is root of this equation
if i let your equation is correct then this is not satisfied
by the way i solve this type question
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