fiora

In your second step,the denominator did not change; you wrote 3 instead of 2 as the numerator in the first frist fraction.And there is something else wrong in your work.

You can do it in this way,

  $${\frac{{\mathtt{2}}}{{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}^{{\mathtt{2}}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{3}}}{\left({{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{9}}\right)}} = {\mathtt{0}}$$

plus 3/(x^2-9) both side

$${\frac{{\mathtt{2}}}{{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}^{{\mathtt{2}}}}} = {\frac{{\mathtt{3}}}{\left({{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{9}}\right)}}$$

bot side times (x+3),

$${\frac{{\mathtt{2}}}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}} = {\frac{{\mathtt{3}}}{\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}}$$

$${\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{6}} = {\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{9}}$$

$${\mathtt{x}} = -{\mathtt{15}}$$

check: 2/(-15+3)^2-3/(x^2-9)=1/72-1/72=0 and x=-15 not equal to 3 or -3.

another way:

$${\frac{{\mathtt{2}}}{{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}^{{\mathtt{2}}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{3}}}{\left({{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{9}}\right)}} = {\mathtt{0}}$$

$${\frac{{\mathtt{2}}}{\left[\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right){\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)\right]}}{\mathtt{\,-\,}}{\frac{{\mathtt{3}}}{\left(\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right){\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}\right)\right)}} = {\mathtt{0}}$$

$${\frac{{\mathtt{2}}{\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}{\left[\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right){\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right){\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}\right)\right]}}{\mathtt{\,-\,}}{\frac{{\mathtt{3}}{\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}{\left[\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right){\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right){\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}\right)\right]}} = {\mathtt{0}}$$

both sides of the equation multiply by (x+3)^2(x-3) (x not equal 3 or -3)

$${\mathtt{2}}{\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}\right){\mathtt{\,-\,}}{\mathtt{3}}{\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right) = {\mathtt{0}}$$ 

$${\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{9}} = {\mathtt{0}}$$

$${\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{15}} = {\mathtt{0}}$$

$${\mathtt{x}} = -{\mathtt{15}}$$

Thank you,Alan.I will fix it.

4*cos^2(pi/12)-2?

cos(2x)=2*cos^2(x)-1

2*cos(2x)=4*cos^2(x)-2

therefor,$${\mathtt{4}}{\mathtt{\,\times\,}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\frac{{\mathtt{\pi}}}{{\mathtt{12}}}}\right)}}^{\,{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}} = {\mathtt{2}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{12}}}}\right)} = {\mathtt{2}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\frac{{\mathtt{\pi}}}{{\mathtt{6}}}}\right)} = {\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}}{{\mathtt{2}}}} = {\sqrt{{\mathtt{3}}}}$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{a}}\right)} = {\frac{{\mathtt{7}}}{{\mathtt{24}}}}$$   $${\frac{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{a}}\right)}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{a}}\right)}}} = {\frac{{\mathtt{7}}}{{\mathtt{24}}}}$$    $$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{a}}\right)} = {\frac{{\mathtt{7}}}{{\mathtt{24}}}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{a}}\right)}$$    $${\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{a}}\right)}}^{\,{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{a}}\right)}}^{\,{\mathtt{2}}} = {\mathtt{1}}$$

substitute sin(a)=7/24*cos（a) into sin^2(a)+cos^2(a）=1

we have 49/576*cos^2(a)+cos^2(a)=1

             625/576*cos^2(a)=1

                      cos^2(a)=576/625

since  tan(a)>0 ,so we have,   

In first  quadrant ,    0<a<pi/2              cos（a)=24/25   sin(a)=7/24*(24/25)=7/25

    In third  quadrant,     pi<a<3pi/2        cos(a)=-24/25   sin(a)=7/24*(-24/25)=-7/25

given that cos(b)=-12/13   pi/2<b<3pi/2

when  pi/2<b0 sin(b)=[1-(-12/13)^2]^(1/2)=5/13

when pi/2<b

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{a}}{\mathtt{\,\small\textbf+\,}}{\mathtt{b}}\right)} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{a}}\right)}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{b}}\right)}{\mathtt{\,\small\textbf+\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{a}}\right)}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{b}}\right)}$$

soloution 1 :when 0<a<pi/2 and pi/2<b<pi

sin(a+b)=7/25*(-12/13)+24/25*(5/13)=36/325

soloution 2 :when 0<a<pi/2 and pi<b<3/2*pi

sin(a+b)=7/25*(-12/13)+24/25*(-5/13)=-204/325

soloution 3:when pi<a<3pi/2 and pi/2<b<pi

sin(a+b)=-7/25*(-12/13)+(-24/25)*(5/13)=-36/325

soloution 4: when pi<a<3pi/2 and pi<b<3/2*pi

sin(a+b)=-7/25*(-12/13)+(-24/25)*(-5/13)=204/325

(edited)

Thank you,Melody!

I usually try to figure out anwsers of my questions before I post it into the forum.I am having fun with doing challenges question like this.

I will not present like "I had done this" in my question,but I will indicatet that  if I dont know how to do my question.So,you could do my  question when you have time.Is that okay? Thank you! 

I had done this before I post it in the forum.I post this because I think you guys could learn something from this question.If you don't,I will not post this kind question anymore.

Ya. Did you learn something from the anwser? 

No,the question given that the measure of angle NOP is 30 degrees.

Indeed ,this is a translation problem.

Acctually,this question was created by chinese.I translate it and post it on the forum.Yes,I am a chinese.

See,

I guess you dont want to do this question anymore ,so I post the anwsers 

Math is an international language.

Okay,Melody.At first I would like to apologize for that.And thank you for spend your time to solve the question.

Now,I have to explain my question.I stated "if" at the begin of each questions.That mean you can not use the given in qustion 1 to solve qustion 2.

But if I did state "if" at the begin of the qustion ,like "$${\mathtt{OM}} = {\sqrt{{\mathtt{5}}}}$$,what is the length of PM?",then you can use it to solve question 2. and I stated if,so.......

Now,if you could only use the given in number two and the title,can you do qustion 2? 

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