Question

Find the value of A(0<=A<=90) when sin^2A-3sinA+2=0

Answer 1

Answer:

Required values of A is 90° .

Step-by-step explanation:

= > sin^2 A - 3sinA + 2 = 0

= > sin^2 A - ( 2 + 1 )sinA + 2 = 0

= > sin^2 A - 2sinA - sinA + 2 = 0

= > sinA( sinA - 2 ) - ( sinA - 2 ) = 0

= > ( sinA - 2 )( sinA - 1 ) = 0

Since their product is 0, one of them must be 0.

If sinA - 2 = 0

sinA = 2 { sine of any angle can't be more than 1 , so sinA ≠ 2 }

If sinA - 1 = 0

sinA = 1

sinA = sin90°

A = 90°

Hence the required values of A is 90° .

Answer 2

Question

Find the value of A for which sin²x - 3sin x + 2 = 0 if 0 ≤ x ≤ 90.

Solution

Given Equation,

$\sf{ {sin}^{2}x - 3sin \: x + 2 = 0 } \\ \\ \hookrightarrow \: \sf{ {sin}^{2}x - 2sin \: x - sin \: x + 2 = 0 } \\ \\ \hookrightarrow \: \sf{sin \: x(sin \: x - 2) - 1(sin \: x - 2) = 0} \\ \\ \hookrightarrow \: \sf{(sin \: x - 2)(sin \: x - 1) = 0} \\ \\ \large{\hookrightarrow \: \boxed{ \boxed{\sf{sin \: x = 2 \: or \: 1}}}}$

sin x = 2 can't be possible because the maximum permissible value of sine function is 1 while the minimum value is -1

Therefore,

$\tt{sin \: x = 1} \\ \\ \longrightarrow \: \tt{sin \: x = sin90{}^{ \circ}} \\ \\ \large{ \longrightarrow \: \boxed{ \boxed{ \tt{x = {90}^{ \circ} }}}}$