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S_A_V [24]
3 years ago
15

Solve the equation for z^3-z^2-12z=0 The solutions are z=?, z=?, and z=?

Mathematics
1 answer:
Assoli18 [71]3 years ago
3 0

Answer:

z is equal to 0,4, and -3

Step-by-step explanation:

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Solve the following equation for 0° ≤ θ < 360°. Use the "^" key on the keyboard to indicate an exponent. For example, sin2x w
never [62]

<u>Answer-</u>

\boxed{\boxed{x=90^{\circ},270^{\circ}}}

<u>Solution-</u>

The given equation-

\Rightarrow 2\sin^2 x-\cos^2 x-2=0

As

\sin^2 x+\cos^2 x=1\ \ \Rightarrow \cos^2 x=1-\sin^2 x

Putting this,

\Rightarrow 2\sin^2 x-(1-\sin^2 x)-2=0

\Rightarrow 2\sin^2 x-1+\sin^2 x-2=0

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

\Rightarrow 3\sin^2 x=3

\Rightarrow \sin^2 x=1

\Rightarrow \sin x=\sqrt1

\Rightarrow \sin x=\pm 1

\Rightarrow \sin x=1,\ \sin x=-1

\Rightarrow x=\sin^{-1}(1),\ x=\sin^{-1}(-1)

\Rightarrow x=90^{\circ}+n360^{\circ},\ x=270^{\circ}+n360^{\circ}

Where n=0, 1, 2, ......

As given 0° ≤ x < 360°, so putting n = 0

\Rightarrow x=90^{\circ},\ x=270^{\circ}


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