Question

solve: 2cos^2 x + 3sinx = 0
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Answer 1

Answer:

2cos^2 x + 3sinx = 0

2 ( 1 - sin^2x ) + 3sinx = 0

2 - 2sin^2x + 3sinx = 0

- 2sin^2x + 3sinx + 2 = 0

let sin x = a

- 2 a ^2 + 3 a + 2 = 0

2 a^2 - 3a - 2 = 0

2 a^2 - 4a + a - 2 = 0

2a ( a - 2 ) + a - 2 = 0

( a - 2 ) ( 2a + 1 ) = 0

a - 2 = 0 or 2a + 1 = 0

a = 2 or a = - 1/2

therefore a = sin x = 2 or - 1/2

value of sin x is always between -1 and 1

so we only take sin x = -1/2

sin x = - 1/2

sin x = sin 7π/6

General solution

when sin x = sin y

x = nπ + ( - 1 ) ^n y ,where n ∈ Z

here

y = 7π/6

therefore

x = nπ + ( - 1 ) ^n 7π/6 where n ∈ Z ( answer )

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