p3 tough question

[crash250]

can someone knows how to solve this equation 3cos2x + 4sin2x = 2 btw 0 and 360 :shock: :shock: :shock:

 

[ahmed t]

take 3cos(2x) to the otherside with2
square both sides
change the sin^2(2x) into cos^2(2x) and solve a quadratic for cos2x
then after that find x

 

[ahmed t]

correct me if im wrong

 

[WellWIshER]

are da ans 59.9 -6.65, 239.8, 173.4??????

i spent 35min on it!!!!

damn i am so nt prepared!!

 

[crash250]

THNX BUDDY

 

[WellWIshER]

r they correct

 

[ahmed t]

i got 8 ansers is that correct crash?

 

[WellWIshER]

wt r they mate?

 

[infinity00]

Nothing Is Tough Buddy. Infact, We Make Things Difficult. You Just Need To Concentrate A Bit...
Do It This Way: First Express The Equation on The LHS In Terms Of RSin(u+a), Where u=2x & a is a constant.
You're Gonna Get 3cos2x + 4sin2x = 5Sin(u+36.87), Where R = 5 And a = 36.87

Now, Put 5Sin(u+36.87) Equal To 2 & Solve For x.

Remember To Replace u with 2x In The Final Calculation..

You'll Get Four Answers....173.36, 59.78, 239.78, 353.36

Hope This Helps Buddy...

 

[WellWIshER]

i did da same but instead i did rcos(u-a)

 

[nealDSA]

@ infinity .. Should We Always Attempt Such Questions In The Manner U Suggested & Is There Any Clue\Hint That Could Be Used To Identify That This Method Must Be Used :fool: :unknown: ???!!!

Thanks In Advance

 

[crash250]

thnx guys..............COMMEN SENCE got it

 

[yasser37]

can someone write this and attach it
I don't get it