Question

if 2sinA= 2+ cosA, find sinA

Answer 1

Answer:

$\huge\underline{ \underline{ \red{your \: \: answer}}}$

Step-by-step explanation:

$\sf\red{2sina = 2 + cosa} \\ \sf\red{ = > {(2sin)a}^{2} = {(2 + cos)a}^{2} } \\ \sf\red{ = > {4sin}^{2}a = 4 + {cos}a^{2} + 4cosa } \\ \sf\red{ = > {4sina}^{2} - 4 - {cos}^{2}a - 4cosa = 0 } \\ \sf\red{ = > 4(1 - {cos}^{2}a ) - 4 - {cos}^{2}a - 4cosa = 0} \\ \sf\red{ = > 4 - {4cos}^{2}a - 4 - {cos}^{2}a - 4cosa = 0} \\ \sf\red{ = > - 5 {cos}^{2}a - 4cosa = 0 } \\ \sf\red{ = >cosa ( - 5cosa +4 ) = 0} \\ \sf\red{ cosa = 0 \: \: \: - 5cosa + 4 = 0} \\ \sf\red{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = > cosa = \frac{ - 4}{ - 5} = \frac{4}{5} } \\ \sf\blue{we \: know \: that \: {sin}^{2}a + {cos}^{2}a = 1 } \\ \sf\pink{ {sin}^{2}a + 0 = 1 = > sina = 1 } \\ \sf\purple{ {sin}^{2}a + { (\frac{4}{5}) }^{2} = 1 } \\ \sf\purple{ {sin}^{2} a + \frac{16}{25} = 1 } \\ \sf\purple{ {sin}^{2}a = 1 - \frac{16}{25} } \\ \sf\purple{ {sin}^{2}a = \frac{9}{25} } \\ \sf\purple{ sina = \sqrt{ \frac{9}{25} } } \\ \sf\purple{sina = \frac{3}{5} } \\ \large\boxed{ \fcolorbox{red}{blue}{hope \: it \: help \: you}}$

Answer 2

Answer:

if 2sinA=2+cosA,

Let SinA = x.

So cosA = (1-x2)1/2.

Hence the equation,

2x = 2 - (1-x2)1/2. Or (1-x2)1/2

= 2 - 2xSquaring,

we have,

4+4x2-8x

= 1-x2.or 5x2 - 8x + 3

= 0. (5x-3)(x-1)

= 0.So x

= 1 or x = 3/5 So SinA = 1 or 3/5.