Question

solve this question:-

4 sin^2 x – 8 cos x + 1 = 0​

Answer 1

Step-by-step explanation:

4 sin ^2 X - 8cos X + 1 = 0

4 ( 1- cos^2 X ) - 8 cos X +1 = 0

4-4 cos^2 X -8 cos X + 1 = 0

- 4 cos ^2 - 8 cos X +5 = 0

4 cos ^2 X + 8 cos X -5 = 0

put cos X = y

4 y^2 + 8y -5 = 0

4y^2 + 10y -2y -5 = 0

2y ( 2y +5) -1( 2y+5) = 0

(2y-1)(2y+5) = 0

put the value of y

(2*cos X -1) ( 2* cos X+5) = 0

2cos X - 1 = 0

2cos X = 1

cos X = 1/2

cos X = cos 60

X = 60°

or

2cos X + 5 = 0

cos X = -5/2

Answer 2

Answer:

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