Question

solve :
sin²x-cosx=1/4 where X belongs to [0,2π]​

Answer 1

Answer:

sin

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

$1 - \cos ^{2} (x) - \cos(x) = 1 \div 4$

$4\cos ^{2} (x) + 4\cos(x) - 3 = 0$

let t=cos^2 (x)

4t^2+4t-3=0

4t^2-2t+6t-3=0

(2t+3)(2t-1)=0

therefore cosx= -3/2 or 1/2

but cosx is not equal to -3/2 as it is not in a range of cosx

therefore cosx=1/2

i. e.

$x = \pi \div 3$

or

$x= 5\pi \div 3$