Question

Find all solutions of the equation in the interval [0, 2pi).

Answer 1

Answer:

x = 0 , π

Step-by-step explanation:

$-4 \sin x = 1 - cos^2 x$

• Rewrite it by using the identity $\sin^2x + \cos^2x = 1$

$=> -4\sin x = \sin^2x$

• Add 4sin x to both the sides.

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

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

• Take sin x common from the expression in L.H.S.

$=> \sin x(\sin x + 4)=0$

Here , we can get two more equations to find x.

1) $\sin x(\sin x + 4)=0$

• Divide both the sides by sin x

$=> \frac{\sin x(\sin x + 4)}{\sin x} = \frac{0}{\sin x}$

$=> \sin x + 4 = 0$

• Substract 4 from both the sides.

$=> \sin x + 4 - 4 = 0 - 4$

$=> \sin x = -4$

$=> x = No \; Solution$

2) $\sin x(\sin x + 4)=0$

• Divide both the sides by (sin x + 4)

$=> \frac{\sin x(\sin x + 4)}{\sin x + 4} = \frac{0}{\sin x + 4}$

$=> \sin x = 0$

$=> x = 0 \; , \pi$ over interval [0 , 2π).