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love history [14]
3 years ago
15

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

Mathematics
1 answer:
mina [271]3 years ago
5 0

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).

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