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sin(3x)=cos(5x)

how to solve this

\(\small{ \begin{array}{rcl} &\text{Formula } \\ &\boxed{~ \begin{array}{rcl} \cos{(A)}-\cos{(B)} &=& -2 \cdot \sin{(\frac{A+B}{2})} \cdot \sin{(\frac{A-B}{2})}\\ \end{array}\\ ~}\\\\ \end{array}\\ \begin{array}{lrcll} & \sin{(3x)}&=& \cos{(5x)} \\ & \sin{(3x)}- \cos{(5x)}&=& 0 \qquad | \qquad \sin{(3x)}=\cos{( \frac{\pi}{2}-3x )} \\ & \cos{( \underbrace{ \frac{\pi}{2}-3x }_{=A} )}- \cos{( \underbrace{ 5x }_{=B} )}&=& 0 \\ & \cos{(\frac{\pi}{2}-3x )}- \cos{( 5x )} &=& -2\cdot \sin{(\frac{\frac{\pi}{2}-3x+5x }{2})} \cdot \sin{(\frac{\frac{\pi}{2}-3x-(5x) }{2})} \qquad \text{Formula}\\ & &=& -2\cdot \sin{(\frac{\frac{\pi}{2}+2x }{2})} \cdot \sin{(\frac{\frac{\pi}{2}-8x }{2})}\\\\ & \mathbf{ \sin{(3x)}- \cos{(5x)} } & \mathbf{=} & \mathbf{ -2\cdot \sin{ ( \frac{\pi}{4}+x) } \cdot \sin{( \frac{\pi}{4}-4x)} = 0 }\\\\ & -2\cdot \sin{ ( \frac{\pi}{4}+x) } \cdot \sin{( \frac{\pi}{4}-4x)} &=& 0 \qquad | \qquad : -2\\ & \underbrace{ \sin{ ( \frac{\pi}{4}+x) } }_{=0} \cdot \underbrace{ \sin{( \frac{\pi}{4}-4x)} }_{=0}&=& 0\\\\ \hline 1. \text{ Solution} & \sin{ ( \frac{\pi}{4}+x ) } &=& 0 \\ & \frac{\pi}{4}+x &=& \arcsin{ ( 0 ) } \pm 2k\cdot \pi \\ & \frac{\pi}{4}+x &=& \pm 2k\cdot \pi \\ & \mathbf{x} &\mathbf{=}&\mathbf{ - \frac{\pi}{4} \pm 2k\cdot \pi } \qquad k \in Z\\ \hline 2. \text{ Solution} & \sin{ ( \pi - ( \frac{ \pi}{4}+x ) ) } &=& 0 \\ & \sin{ ( \frac{3 \pi}{4}-x ) } &=& 0 \\ & \frac{3 \pi}{4}-x &=& \arcsin{ ( 0 ) } \pm 2k\cdot \pi \\ & \frac{3 \pi}{4}-x &=& \pm 2k\cdot \pi \\ & \mathbf{x} &\mathbf{=}& \mathbf{\frac{3 \pi}{4} \pm 2k\cdot \pi } \qquad k \in Z\\ \hline 3. \text{ Solution} & \sin{ ( \frac{\pi}{4}-4x ) } &=& 0\\ & \frac{\pi}{4}-4x &=& \arcsin{ ( 0 ) } \pm 2k\cdot \pi \\ & \frac{\pi}{4}-4x &=& \pm 2k\cdot \pi \\ & 4x &=& \frac{\pi}{4} \pm 2k\cdot \pi \\ & \mathbf{x} &\mathbf{=}& \mathbf{\frac{\pi}{16} \pm k\cdot \frac{\pi}{2} } \qquad k \in Z\\ \hline 4. \text{ Solution} & \sin{ ( \pi - ( \frac{ \pi}{4}-4x ) ) } &=& 0\\ & \sin{ ( \frac{3\pi}{4}+4x ) } &=& 0\\ & \frac{3\pi}{4}+4x &=& \arcsin{ ( 0 ) } \pm 2k\cdot \pi \\ & \frac{3\pi}{4}+4x &=& \pm 2k\cdot \pi \\ & 4x &=& -\frac{3\pi}{4} \pm 2k\cdot \pi\\ & \mathbf{x} &\mathbf{=}& \mathbf{ -\frac{3\pi}{16} \pm k\cdot \frac{\pi}{2} } \qquad k \in Z\\ \end{array} }\)

Wed  18/11/15

1)  Quarters and nickels

http://web2.0calc.com/questions/money-and-percents#r1

2)  Omi57's pic should bring a smile to everybody's face :))     Thanks Omi67

http://web2.0calc.com/questions/a-tree-casts-a-25-foot-shadow-on-sunny-day-if-the-angle-of-elevation-from-the-tip-of-the-shadow-to-the-top-of-the-tree-is-32-degree-what-is-the-height-of#r3

3) Complex plane

http://web2.0calc.com/questions/complex-plane#r3

Ok so mine is correct for the question that I answered.

A question to the asker.

Why would you write exp if you meant 'e' ?

They are not the same thing at all !!

Thank you Alan :)

@@ What is Happening?  [Wrap4]   Tues 17/11/15   Sydney, Australia Time 8:32pm  ♪ ♫

Good evening to you all,

We have had some wonderful answers from Meghara, CPhill, Heureka, DoctorNewton, Omi67, LordEnedar, Rom, Alan and TMga2yb.  A big thank you to each of you.  

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Interest Posts:

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1) Calling our comedian TROLLS to return.    Continued.  

     Please, this place needs an injection of comedy.    

Divide and simplify.

t^2+8t/t^2+3t-40 ÷ t/t+5

*No brakets*

Thanks!!!

I assume ( t^2+8t ) / ( t^2+3t-40 ) ÷ t/ (t+5)  

\(\begin{array}{rcl} \dfrac{ \dfrac{( t^2+8t )} { ( t^2+3t-40 ) } } { \dfrac{ t} { (t+5) } } &=& \dfrac{ \dfrac{( t^2+8t )} { ( t+8)(t-5 ) } } { \dfrac{ t} { (t+5) } } \\\\ &=& \dfrac{( t^2+8t )} { ( t+8)(t-5 ) } \cdot \dfrac{ (t+5) }{ t} \\\\ &=& \dfrac{( t^2+8t )} { ( t+8)(t ) } \cdot \dfrac{ (t+5) }{ (t-5) } \\\\ &=& \dfrac{( t^2+8t )} { ( t^2+8t ) } \cdot \dfrac{ (t+5) }{ (t-5)} \\\\ &=& \dfrac{ (t+5) }{ (t-5)} \\\\ \end{array}\)

Need 10exp(-0.17x) to be interpreted as 10*e^(-0.17x) rather than 10^(-0.17x).

Alan, could you please what is wrong with my logic ?

That is easy, anything divided by 1 stays the same.      

-i /1 = -i

Here are graphs showing two solutions and the corresponding numerical values obtained using Newton-Raphson:

y^2-81/25y+225 ÷ y-9/45

(y^2-81)/(25y+225) ÷ (y-9)/45

if this is what you mean it would look like this - The fraction line replaces the brackets.

\(\frac{y^2-81}{25y+225} ÷ \frac{y-9}{45}\)

Is that what your question looks like?

--------------------

ok I will assume that it is.

\(\frac{y^2-81}{25y+225} ÷ \frac{y-9}{45}\\ =\frac{y^2-81}{25y+225} \times \frac{45}{y-9}\\ =\frac{(y-9)(y+9)}{25(y+9)} \times \frac{45}{y-9}\\ =\frac{(y+9)}{25(y+9)} \times \frac{45}{1}\\ =\frac{1}{25} \times \frac{45}{1}\\ =\frac{1}{5} \times \frac{9}{1}\\ =\frac{9}{5} \\ =1\frac{4}{5} \\\)

-7x-3(9-7x) if x=-1

   -7x        -3  (9   - 7 x)    if x=-1

= -7*-1    -3*(9  -  7*-1) 

= +7       -3*(9 - -  7   )

= 7         -3*(9 +  7  )

= 7        -3* 16

= 7    -   48

= -41

I  x+y=3

II x*y=1.25

I  x=3-y

II (3-y)*y=1.25

II -y^2+3y-1.25=0

II y^2 -3y+1.25=0

a=1

b=-3

c=1.25

y1,2=  (3+/- sqrt(9-4*1.25))/2

y1,2= (3+/-2)/2

y1=2.5  => x=3-2.5=0.5

y2=0.5 => x=3-0.5=2.5

The numbers : 0.5, 2.5

It does not come with brakets in my homework,sorry

Could you put some brackets in please ?

simplified radical form of √208

\(\sqrt{208}=\sqrt{4*4*13}=4\sqrt{13}\)

17/30=0.5666666667

0.566666667*100=56.66666667

Answer:  56.66666667%

Divide and simplify.

3x-15/121 ÷ x-5/33

\(\begin{array}{rcl} \dfrac{ \frac{3x-15}{121} }{ \frac{x-5}{33} } &=& \dfrac{ \frac{3(x-5)}{121} }{ \frac{x-5}{33} } \\\\ &=& \frac{3(x-5)}{121} \cdot \frac{33}{(x-5)} \\\\ &=& \frac{3\cdot 33}{121} \\\\ &=& \frac{99}{121} \\\\ &=& \frac{9}{11} \end{array}\)

0.6=3/30+x

\(0.6=\frac{3}{30}+x\\ 0.6=\frac{1}{10}+x\\ 0.6=0.1+x\\ 0.6-0.1=0.1-0.1+x\\ 0.5=x\\ x=0.5 \)

I needed a laugh,  thanks Rom  

Mmmm   What is the question?

What is between the x and the 10?

x 10exp(−0.17x)=1 what is the value of x

x ? 10^(-0.17x) = 1

Is it meant to be multiply??

\(x \times 10^{-0.17x} = 1\\ x \times \frac{10^{-0.17x}}{1} = 1\\ x \times \frac{1} {10^{0.17x}}= 1\\ x=10^{0.17x} \\ log x=0.17x\)

I am going to cheat a bit and get a graphical solution

I am graphing y=log(x)  and  y=0.17x

Looks to me like there is no solution since the 2 graphs don't intersect.

Can someone check this please?

bout tree fitty

Divide and simplify the following expression.

 

x^2-49/x ÷ x+7/x-7

 

\(\dfrac{\frac{x^2-49}{x}}{\frac{x+7}{x-7}}=\dfrac{x^2-49}{x}\dfrac{x-7}{x+7}=\) 

\(\dfrac{(x-7)(x+7)(x-7)}{x(x+7)}=\dfrac{(x-7)^2}{x}\)

Thanks Rom   

Hi All how do you rationalise and symplify 2 root 3 over root 7

\(\dfrac {2\sqrt{3}}{\sqrt{7}} = \dfrac {2\sqrt{3}}{\sqrt{7}} \dfrac{\sqrt{7}}{\sqrt{7}} = \dfrac {2\sqrt{21}}{7}\)