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RideAnS [48]
3 years ago
6

Please answer this question !! BRAINLIEST !!

Mathematics
1 answer:
Misha Larkins [42]3 years ago
8 0

Cos inverse0.7813

=38.6°

B

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The difference (subtraction) of a number (x) squared (²) and the number (x) is (=) one less than (- 1) four times (4 ·) the number (x)

x² - x = 4x - 1
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Is A’B’C’ congruent to ABC explain why or why not
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Is there a picture?

Step-by-step explanation:

please add a picture. You can also see if the transition is congruent by counting

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Which of the following statements is true?
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B) Aluminum and steel are good conductors of electricity.

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2 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
I really need help please
Lana71 [14]
Oh um... well let's see... okay I do NOT know this. SO sorry. Hope you get an answer soon.
8 0
2 years ago
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