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Andrei [34K]
3 years ago
5

|2s|+19=13 Help me please

Mathematics
1 answer:
Mnenie [13.5K]3 years ago
4 0

 \textsf{Hey there!}

\mathsf{|2s| + 19= 13}\\\textsf{Solve both sides by -19}\\\mathsf{|2s|+19+ (-19)=13+(-19)}\\\mathsf{Cancel\ out:\ 19 + (-19) \  because\ that\ gives\ you\ 0}\\\mathsf{Keep: 13 + (-19)\ because\ that\  helps\ us\ solve\ for\ our\ answer}\\\mathsf{13 + (-19) = -6}\\\mathsf{Solve\ for\ absolute\ value: |2s| = -6}\\\mathsf{Any\ absolute\ values\ CANNOT\ be\ less\ than\ 0}\\\mathsf{\boxed{\boxed{\mathsf{Thus, \ this\ make\ your\ answer\ is: \boxed{\boxed{\bf{No\ solution}}}}}}}\checkmark

\textsf{Good luck on your assignment and enjoy your day!}

~\frak{LoveYourselfFirst:)}

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If sin θ = 1/2 , then find the value of (sin 3 θ)/(1+ cos 2 θ)<br>​
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<u>EXPLANATION</u><u>:</u>

Given that

sin θ = 1/2

We know that

sin 3θ = 3 sin θ - 4 sin³ θ

⇛sin 3θ = 3(1/2)-4(1/2)³

⇛sin 3θ = (3/2)-4(1/8)

⇛sin 3θ = (3/2)-(4/8)

⇛sin 3θ = (3/2)-(1/2)

⇛sin 3θ = (3-1)/2

⇛sin 3θ = 2/2

⇛sin 3θ = 1

and

cos 2θ = cos² θ - sin² θ

⇛cos 2θ = 1 - sin² θ - sin² θ

⇛cos 2θ = 1 - 2 sin² θ

Now,

cos 2θ = 1-2(1/2)²

⇛cos 2θ = 1-2(1/4)

⇛cos 2θ = 1-(2/4)

⇛cos 2θ = 1-(1/2)

⇛cos 2θ = (2-1)/2

⇛cos 2θ = 1/2

Now,

The value of sin 3θ /(1+cos 2θ

⇛1/{1+(1/2)}

⇛1/{(2+1)/2}

⇛1/(3/2)

⇛1×(2/3)

⇛(1×2)/3

⇛2/3

<u>Answer</u> : Hence, the req value of sin 3θ /(1+cos 2θ) is 2/3.

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