Answer:
x = 0 , π
Step-by-step explanation:
[tex]-4 \sin x = 1 - cos^2 x[/tex]
- Rewrite it by using the identity [tex]\sin^2x + \cos^2x = 1[/tex]
[tex]=> -4\sin x = \sin^2x[/tex]
- Add 4sin x to both the sides.
[tex]=> -4\sin x + 4\sin x = sin^2x + 4\sin x[/tex]
[tex]=> \sin^2x + 4\sin x = 0[/tex]
- Take sin x common from the expression in L.H.S.
[tex]=> \sin x(\sin x + 4)=0[/tex]
Here , we can get two more equations to find x.
1) [tex]\sin x(\sin x + 4)=0[/tex]
- Divide both the sides by sin x
[tex]=> \frac{\sin x(\sin x + 4)}{\sin x} = \frac{0}{\sin x}[/tex]
[tex]=> \sin x + 4 = 0[/tex]
- Substract 4 from both the sides.
[tex]=> \sin x + 4 - 4 = 0 - 4[/tex]
[tex]=> \sin x = -4[/tex]
[tex]=> x = No \; Solution[/tex]
2) [tex]\sin x(\sin x + 4)=0[/tex]
- Divide both the sides by (sin x + 4)
[tex]=> \frac{\sin x(\sin x + 4)}{\sin x + 4} = \frac{0}{\sin x + 4}[/tex]
[tex]=> \sin x = 0[/tex]
[tex]=> x = 0 \; , \pi[/tex] over interval [0 , 2π).
