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Angelina_Jolie [31]

2 years ago

8

PLEASE HELP ASAP!!! Normal people- me- I WILL MARK FIRST ANSWER BRAINLIEST BUT IT HAS TO BE AN ACTUAL ANSWER (the question is in

the image)

1 answer:

Tasya [4]2 years ago

5 0

Answer:

It would be B

Step-by-step explanation:

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matrenka [14]

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the correct answer should be 109

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Kaylis [27]

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3 years ago

Find all solutions of the equation: 2cos^2x-cosx=1

Art [367]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3166243

——————————

Solve the trigonometric equation:

$\mathsf{2\,cos^2\,x-cos\,x=1}\\\\ \mathsf{2\,cos^2\,x-cos\,x-1=0}$

Make a substitution:

$\mathsf{cos\,x=t\qquad (-1\le t\le 1)}$

and the equation becomes

$\mathsf{2t^2-t-1=0}$

Rewrite conveniently – t as + t – 2t, and then factor the left-hand side by grouping:

$\mathsf{2t^2+t-2t-1=0}\\\\ \mathsf{t\cdot (2t+1)-1\cdot (2t+1)=0}$

Factor out 2t + 1:

$\mathsf{(2t+1)\cdot (t-1)=0}\\\\ \begin{array}{rcl} \mathsf{2t+1=0}&~\textsf{ or }~&\mathsf{t-1=0}\\\\ \mathsf{2t=1}&~\textsf{ or }~&\mathsf{t=1}\\\\ \mathsf{t=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{t=1} \end{array}$

Substitute back for t = cos x:

$\begin{array}{rcl}\mathsf{cos\,x=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{cos\,x=1}\\\\ \mathsf{cos\,x=cos\,60^\circ}&~\textsf{ or }~&\mathsf{cos\,x=cos\,0} \end{array}$

Therefore,

$\begin{array}{rcl} \mathsf{x=\pm\,60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{cos\,x=0+k\cdot 360^\circ} \end{array}$

where k is an integer.

Solution set:

$\mathsf{S=\left\{x\in\mathbb{R}:~~x=-\,60^\circ+k\cdot 360^\circ~~or~~x=60^\circ+k\cdot 360^\circ~~or~~x=k\cdot 360^\circ,~~k\in\mathbb{Z}\right\}}$

I hope this helps. =)

3 0

3 years ago

Which of the following stanements shows the inverse property of addition?

Nikitich [7]

Answer:

$0\cdot \:y+\left(-0\cdot \:y\right)=0$ the statement shows the inverse property of addition as the sum of a number and its inverse is zero.

Step-by-step explanation:

As we know that Inverse Property of Addition states if you add any number to its opposite, the result will be zero.

For example, 13 + (-13) = 0 shows that -13 is the additive inverse of 13.

In other words, the sum of a number and its inverse is zero.

Now checking from the available options,

$0y+\left(-0y\right)$

$\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a$

$=0\cdot \:y-0\cdot \:y$

$\mathrm{Add\:similar\:elements:}\:0\cdot \:y-0\cdot \:y=0$

$=0$

Thus,

$0\cdot \:y+\left(-0\cdot \:y\right)=0$

Therefore, $0\cdot \:y+\left(-0\cdot \:y\right)=0$ the statement shows the inverse property of addition as the sum of a number and its inverse is zero.

4 0

3 years ago