Question

If 2sinβ + 3 cos β =2 , prove that ( 3 sin β – 2cos β) = ± 3​

Answer 1

Answer:

Let z1=cosα+isinα=eiα, z2=cosβ+isinβ=eiβ and z3=cosγ+isinγ=eiγ

Now z1+2z2+3z2=(cosα+2cosβ+3cosγ)+i(sinα+2sinβ+3sinγ)=0

Now using the fact: If a+b+c=0, then a3+b3+c3=3abc

(z1)3+(2z2)3+(3z3)3=3(z1)(2z2)(3z3)

⇒e3iα+8e3iβ+27e3iγ=18ei(α+β+γ)=18

Answer 2

Correct option is

B

0

Let z1=cosα+isinα=eiα, z2=cosβ+isinβ=eiβ and z3=cosγ+isinγ=eiγ

Now z1+2z2+3z2=(cosα+2cosβ+3cosγ)+i(sinα+2sinβ+3sinγ)=0

Now using the fact: If a+b+c=0, then a3+b3+c3=3abc

(z1)3+(2z2)3+(3z3)3=3(z1)(2z2)(3z3)

⇒e3iα+8e3iβ+27e3iγ=18e