Solving a trigonometric equation

[chwala]

Find the value of x between 0 degrees and 360 degree satisfying the equation
10sin^2x+ 10sin x cos x - cos^2x = 2 this is how i have attempted.....
10 sin^x+ 10sin2x/2 - cos^x = 2
I used the property sin 2x = 2 sin x cos x and substituted sin x cos x with sin 2x/2 giving me....
11sin^2x + 5sin 2x -3 =0 on reaching here i got stuck...kindly advise

 

[pasmith]

Use the identity \sin^2 x = \frac12(1 - \cos 2x).

 

[chwala]

is this the way to go bro...

11(0.5(1-cos 2x)+5 sin 2x - 3 =0

 

[Char. Limit]

Honestly, this doesn't look easy in any way, if I'm completely honest. I did some checking on Wolfram-Alpha to see what needs to be done, and it looks like it needs to be somehow written in terms of the tangent function... which I suppose you could do by dividing the whole thing by cos(2x) and then rewriting the sec(2x) that comes up in terms of the tangent function.

Not that that's going to be easy, but it definitely looks like the way to go.

 

[chwala]

i need more insight on this any help or clue, Char.Limit i still don't understand.....

 

[pasmith]

Honestly, this doesn't look easy in any way, if I'm completely honest. I did some checking on Wolfram-Alpha to see what needs to be done, and it looks like it needs to be somehow written in terms of the tangent function... which I suppose you could do by dividing the whole thing by cos(2x) and then rewriting the sec(2x) that comes up in terms of the tangent function.

Not that that's going to be easy, but it definitely looks like the way to go.

Have you not had to solve problems of the form <br /> A \cos t + B \sin t = C before? This is basic trig: one uses the identity <br /> R\cos(t - \alpha) = R\cos\alpha \cos t + R\sin \alpha \sin t. Setting this equal to A \cos t + B \sin t
yields R \cos \alpha = A \\ R \sin \alpha = B which are easily solved for R and \tan \alpha, and then R \cos(t - \alpha) = C is easily solved for t.

 

[Char. Limit]

Huh. That's actually not any trig I ever learned... very good to know, though. My bad, pas, that's definitely the way to go.

 

[Curious3141]

Find the value of x between 0 degrees and 360 degree satisfying the equation
10sin^2x+ 10sin x cos x - cos^2x = 2 this is how i have attempted.....
10 sin^x+ 10sin2x/2 - cos^x = 2
I used the property sin 2x = 2 sin x cos x and substituted sin x cos x with sin 2x/2 giving me....
11sin^2x + 5sin 2x -3 =0 on reaching here i got stuck...kindly advise

Express everything in terms of trig ratios of twice the angle using the double angle formulae:

\sin 2x = 2\sin x \cos x

and

\cos 2x = 2\cos^2 x - 1 = 1 - 2\sin^2 x

(you'll need both forms of the latter identity).

Then, letting y = 2x (just for clarity), you can transform the equation into the form

A \sin y + B \cos y = C

after which you can proceed as pasmith advised.

 

[chwala]

thanks i attempted this as follows
-11/2 cos 2x + 5 sin 2x = -5/2
-11 cos 2x + 10sin 2x=-5 and letting y=2x
-11 cos y+10sin y= -5
now this is in the form Acos t + Bsin t = c...from here and using the identity given by pasmith
Rcos α= -11 am i on the correct path thanks again curious3141,char.limit and pasmith.
i also need advise on how i can post using mathematical language/symbols like the ones used by pasmith....