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Leona [35]
4 years ago
9

Let t = cosA + sinA find values of t if 2 - sin2A = 2(cosA + sinA)

Mathematics
1 answer:
Zinaida [17]4 years ago
4 0

2 - \sin(2A) = 2(\cos A + \sin A)\\\\2-2\sin A\cos A=2\cos A+2\sin A\\\\2\sin A\cos A+2\cos A+2\sin A-2=0\\\\1+2\sin A\cos A+2\cos A+2\sin A-3=0\\\\\sin^2 A+\cos^2 A+2\sin A\cos A+2\cos A+2\sin A-3=0\\\\(\sin A+\cos A)^2+2\cos A+2\sin A-3=0\\\\(\sin A+\cos A)^2+2(\cos A+\sin A)-3=0\\\\\\t^2+2t-3=0\\\\t^2+2t+1-4=0\\\\(t+1)^2=4\\\\t+1=2\qquad\vee \qquad t+1=-2\\\\\\\boxed{t=1\qquad\vee\qquad t=-3}

When A\in\mathbb{R}

t=\cos A+\sin A\in[-\sqrt{2};\sqrt{2}]\approx[-1.4142;1.4142]

and there is only one answer t = 1.

For A\in\mathbb{C} both values are correct.

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1. \: 3i

2. option D

3. option C

4. option D

5. option C

6. option B

7. option C

8. option D

9. option C

10. option C

Step-by-step explanation:

<h2>1. \:  \sqrt{ - 9}</h2>

\sqrt{ - 1(9)}

\sqrt{ - 1}  \times  \sqrt{  9}

i \times  \sqrt{9}

i \times  \sqrt{ {3}^{2} }

i \times 3

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<h2>2. \:  \sqrt{ - 8}</h2>

\sqrt{ - 1(8)}

\sqrt{ - 1}  \times  \sqrt{8}

i \times  \sqrt{8}

i \times  \sqrt{ {2}^{2} \times 2 }

2i \sqrt{2}

<h2>3. \:  \sqrt{ - 80}</h2>

\sqrt{ - 1}  \times  \sqrt{80}

i \times  \sqrt{80}

4i \:  \sqrt{5}

<h2>4. \:  \sqrt{ - 75}</h2>

\sqrt{ - 1}  \times   \sqrt{75}

i \times  \sqrt{75}

i \times  \sqrt{ {5}^{2} \times 3 }

5i \sqrt{3}

<h2>5. \:  \sqrt{ - 72}</h2>

\sqrt{ - 1}  \times  \sqrt{72}

i \times  \sqrt{72}

i \times ( {6}^{2}  \times 2)

6i \sqrt{2}

<h2>6.  \sqrt{ - 20}</h2>

\sqrt{ - 1}  \times  \sqrt{20}

i \times  \sqrt{20}

i \times  \sqrt{ {2}^{2}  \times 5}

2i \sqrt{5}

<h2>7. \:  \sqrt{ - 27}</h2>

\sqrt{ - 1}  \times  \sqrt{27}

i \times  \sqrt{27}

i \times  \sqrt{ {3}^{2}  \times 3}

3i \sqrt{3}

<h2>8. \:  \sqrt{ - 12}</h2>

\sqrt{ - 1 \times 12}

i \times  \sqrt{12}

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2i \sqrt{3}

<h2>9. \:  \sqrt{ - 125}</h2>

\sqrt{ - 1}  \times  \sqrt{125}

i \times  \sqrt{ {5}^{2} \times 5 }

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<h2>10. \:  \sqrt{ - 180}</h2>

\sqrt{ - 1}  \times  \sqrt{180}

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6i \sqrt{5}

<h3>Hope it is helpful...</h3>
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