Question

Find cosA, if 2sin²A+7cosA=5

Answer 1

Step-by-step explanation:

2 sin ^2A + 7 cos A = 5

7cos A = 5- 2 sin ^2A

cos A = (5 - 2 sin ^2A)/7

hope it helps you.........

Answer 2

Answer:

We have,

$2 {sin}^{2} a \: + \: 7cosa \: = \: 5$

so, this implies,

$2(1 - {cos}^{2} a) \: + \: 7 cosa \: = \: 5 \: \: \: \: \: \: \: \: \: (as \: {sin}^{2} a \: = \: 1 - {cos}^{2} a)$

then,

$2 - 2 {cos}^{2} a \: + \: 7cosa \: - \: 5 \: = \: 0$

thus,

$2 {cos}^{2} a \: - 7cosa \: + \: 3 \: = \: 0$

and, by middle term splitting,

$2 {cos}^{2} a \: - 6cosa \: - \: cosa \: + \: 3 \: = \: 0$

so, we get,

$2cosa(cosa \: - \: 3) \: - 1(cosa \: - \: 3) \: = \: 0$

and, it implies,

$(cosa \: - \: 3)(2cosa \: - \: 1) \: = \: 0$

therefore,

$either \: cosa \: = \: 3 \: or \: cosa \: = \: \frac{1}{2}$

but, since

$- 1 \leqslant cosa \leqslant 1 \:$

hence,

$cosa \: = \: \frac{1}{2} \: \: \: \: \: \: \: \: \: \: \: \: (as \: cosa \: = \: 3 \: not \: possible)$