Question

find the general solution of the equation sin^2A-2cosA+1/4=0​

Answer 1

Step-by-step explanation:

Given :-

Sin² A - 2 Cos A +( 1/4 ) = 0

To find :-

Find the general solution of the equation?

Solution :-

Given equation is

Sin² A - 2 Cos A +( 1/4 ) = 0

=> (4 Sin² A - 8 Cos A + 1)/4 = 0

=> 4 Sin² A - 8 Cos A + 1 = 0×4

=> 4 Sin² A - 8 Cos A + 1 = 0

We know that

Sin² A + Cos² A = 1

=> Sin² A = 1 - Cos² A

=> 4(1 - Cos² A ) - 8 Cos A + 1 = 0

=> 4 - 4 Cos² A - 8 Cos A + 1 = 0

=> - 4 Cos² A - 8 Cos A + 5 = 0

=> -( 4 Cos² A + 8 Cos A - 5) = 0

=> 4 Cos² A + 8 Cos A - 5 = 0

=> 4 Cos² A -2 Cos A + 10 Cos A - 5 = 0

=> 2 Cos A ( 2 Cos A - 1) + 5 ( 2 Cos A - 1 ) = 0

=> (2 Cos A - 1) ( 2 Cos A + 5 ) = 0

=> 2 Cos A - 1 = 0 or 2 Cos A + 5 = 0

=> 2 Cos A = 1 or 2 Cos A = -5

=> Cos A = 1/2 or Cos A = -5/2

=> Cos A can not be -5/2

=> Cos A = 1/2

=> Cos A = Cos 60°

=> A = 60° or π/3

Therefore, A = 60° or π/3

Answer:-

The general solution for the given equation is 60° or π/3

Check :-

If A = 60° then LHS of the given equation

Sin² A - 2 Cos A +( 1/4 )

=> Sin² 60° -2 Cos 60° +(1/4)

=> (√3/2)²- 2(1/2) + (1/4)

=> (3/4)-(2/2)+(1/4)

=> (3/4)-(1)+(1/4)

=> (3+1)/4 - (1)

=> (4/4)-1

=> 1-1

=> 0

=> RHS

LHS = RHS is true for A = 60°

Used formulae:-

• Sin² A + Cos² A = 1

• Sin 30° = 1/2

• Cos 60° = 1/2

• Sin 60° = √3/2

• π = 180°