sin theta tan^2 theta (2sin theta + 3) + tan^2 theta = 0

Question

logs edexcel c2 iygb paper z question 6

Answer 1

let θ=x

sinxtan^2x(2sinx+3)+tan^2x=0

factor

tan^2x([sinx][2sinx+3]+1)=0

tan^2x(2sin^2x+3sinx+1)=0

tan^2x(2sinx+1)(sinx+1)=0

tan^2x=0

x=0º and x=180º

2sinx+1=0

2sinx=-1

sinx=-1/2

x=11∏/6 or 330º

sinx+1=0

sinx=-1

x=270º

 

Answer 2

sin(x) tan²x [2sin(x) + 3)] + tan²x = 0

 

tan²x {sin(x) [2sin(x) + 3] + 1} = 0

 

tan²x [2sin²x + 3sin(x) + 1] = 0

 

Two sets of solutions:

 

(i)

 

tan²x = 0 ⇒ x = 0 or π

 

(ii)

 

2sin²x + 3sin(x) + 1 = 0

 

Let y = sin(x)

 

2y² + 3y + 1 = 0 

 

Solve this quadratic equation.

 

y = -1/2 or -1

 

sin(x) = -1/2 or -1

 

I assume you want solutions between 0 and 2π.

 

sin(x) = -1/2 ⇒ x = 7π/6 or 11π/6

 

sin(x) = -1 ⇒ x = 3π/2, but tan(3π/2) is not defined, so discard this one.

 

Final solution set:  {0, π, 7π/6, 11π/6}

 

These are the same angles as in Arthur's solution, but in radians instead of degrees.