Question

General solution of 2cos²x - 3 sin x = 0

Answer 1

2cos^2x-3sinx=0

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

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

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

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

=>sinx=1/2 and -2
but sinx not exist except €[-1,1]
so,
sinx=1/2 only value
now,
sinx=sinπ/6
x=nπ+(-1)^nπ/6

where n is integers

Answer 2

Answer:

$x=n\pi+(-1)^{\frac{n\pi}{6}}$

Step-by-step explanation:

Given : Expression $2\cos^2x-3\sin x=0$

To find : The general solution of the expression?

Solution :

Expression $2\cos^2x-3\sin x=0$

$2(1-\sin^2x)-3\sin x=0$

$2-2\sin^2x-3\sin x=0$

$2\sin^2x+3\sin x-2=0$

Applying middle term split,

$2\sin^2x+4\sin x-\sin x-2=0$

$2\sin x(\sin x+2)-1(\sin x+2)=0$

$(\sin x+2)(2\sin x-1)=0$

$(\sin x+2)=0,(2\sin x-1)=0$

i.e. Either $\sin x+2=0$

$\sin x=-2$

Which is not possible as value of sin lies between 1 and -1.

So, $2\sin x-1=0$

$\sin x=\frac{1}{2}$

$\sin x=\sin (\frac{\pi}{6})$

General solution is

$x=n\pi+(-1)^{\frac{n\pi}{6}}$