Question

2cos^2x + 3sinx = 0​

Answer 1

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

2cos²x + 3sinx = 0

2(1 - sin²x) + 3sinx = 0

2sin²x - 3sinx - 2 = 0

2sin²x - 4sinx + sinx - 2 = 0

2sinx(sinx - 2) + (sinx - 2) = 0

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

(sinx - 2) = 0

Or,

(2sinx + 1) = 0

sinx = 2

Or,

2sinx + 1 = 0

Now,

2sinx + 1 = 0

${\boxed{\sf\:{sinx=-\dfrac{1}{2}}}}$

${\boxed{\sf\:{-sin\dfrac{\pi}{6}}}}$

${\boxed{\sf\:{sin(\pi+\dfrac{\pi}{6})}}}$

${\boxed{\sf\:{sin\dfrac{7\pi}{6}}}}$

${\boxed{\sf\:{sinx=sin\dfrac{7\pi}{6}}}}$

$\Large{\boxed{\sf\:{x=[n\pi+(-1)^{n}\times\dfrac{7\pi}{6}]}}}$

Answer 2

Question :-- solve 2cos²x + 3sinx = 0

Solution :--

→ 2cos²x + 3sinx = 0

Putting cos²x = (1-sin²x) we get,

→ 2(1-sin²x) + 3sinx = 0

→ 2 - 2sin²x + 3sinx = 0

→ 2sin²x - 3sinx -2 = 0

Splitting the Middle term now,

→ 2sin²x - 4sinx + sinx - 2 = 0

→ 2sinx(sinx-2) +1(sinx -2) = 0

→ (sinx -2)(2sinx+1) = 0

Now,

if sinx-2 = 0

→ sinx = 2. That is not possible as −1 ≤ sin x ≤ 1.

So,

→ 2sinx + 1 = 0

→ 2sinx = (-1)

→ 2sinx = (-1/2)

→ 2sinx = -π/6

____________________________

Now, we know that value of sin is negative in Third and Fourth Quadrant ..

So,

→ Third Quadrant = sin(π + π/6) = sin(7π/6)

→ x = 2nπ + 7π/6

And ,

→ Fourth Quadrant = sin(2π-π/6) = sin(11π/6)

→ x = 2nπ + 11π/6