Question

sin²a-cosa=1/4. please solve it​

Answer 1

$\underline{ \tt \: question}$

$\tt \: solve \: to \: find \: \alpha$

$\tt \: {sin}^{2} \alpha - cos \: \alpha = \frac{1}{4}$

$\underline{ \tt \: solution}$

we have,

${sin}^{2} \alpha - cos \: \alpha = \frac{1}{4}$

(multiplying by 4 both sides)

$4 {sin}^{2} \alpha - 4 \: cos \: \alpha =1$

(using identity )

$(\small{1 - {cos}^{2} \alpha = {sin}^{2} \alpha })$

we will get,

$4( {1 - {cos}^{2} \alpha }) - 4cos \: \alpha = 1 \\ \\ 4 - 4 {cos}^{2} \alpha - 4 \: cos \: \alpha = 1 \\ \\ - 4 {cos}^{2} \alpha - 4cos \: \alpha + 3 = 0$

(multiplying by -1 both sides)

$4 {cos}^{2} \: \alpha + 4cos \: \alpha - 3 = 0$

let us take

$cos \: \alpha = x$

we will get a quadratic polynomial in x variable

$4 {x}^{2} + 4x - 3 = 0 \\ \\ 4 {x}^{2} + 6x - 2x - 3 = 0 \\ \\ 2x(2x + 3) - 1(2x + 3) = 0 \\ \\ (2x - 1)(2x + 3) = 0 \\ \\ x = \frac{1}{2} \: and \: x = \frac{ - 3}{2}$

it means we get two values of

$cos \: \alpha$

but as

$cos\alpha \: cannot \: be \: smaller \: than \: - 1 \: \\ hence \: the \: possible \: value \: we \: get \: is \\ \\ cos \: \alpha = \frac{1}{2}$

also,

cos 60° = 1 / 2

hence,

$\boxed{ \alpha = 60\degree}$