"Solve the equation3sin2x−3cos⁡x−6sin⁡x+2sin⁡2x+3=0" solved and explained

Potherat8mi

Answered question

2022-03-01

Solve the equation

3 \sin^{2} x - 3 \cos x - 6 \sin x + 2 \sin 2 x + 3 = 0

Jenson Mcculloch

Beginner2022-03-02Added 9 answers

\Rightarrow 3 (\sin^{2} x - 2 \sin x + 1) + 2 \sin 2 x - 3 \cos x = 0

\Rightarrow 3 {(1 - \sin x)}^{2} + \cos x (4 \sin x - 3) = 0

We know,

\sin x = \frac{2 t}{1 + t^{2}}, \cos 2 x = \frac{1 - t^{2}}{1 + t^{2}}

where

t = \tan \frac{x}{2}

So,

1 - \sin x = \frac{{(1 - t)}^{2}}{1 + t^{2}}, 4 \sin x - 3 = 4 \frac{2 t}{1 + t^{2}} - 3 = - \frac{3 t^{2} - 8 t + 3}{1 + t^{2}} = 0

If t is fine

3 {(1 - t)}^{4} - (1 - t) (1 + t) (3 t^{2} - 8 t + 3) = 0

Or,

(1 - t) [3 {(1 - t)}^{3} - (1 + t) (3 t^{3} - 8 t + 3)] = 0

t = 1, \tan \frac{x}{2} = 1 \Rightarrow \frac{x}{2} = n π + \frac{π}{4} \Rightarrow x = 2 n π + \frac{π}{2}

where n is any integer.
Else

2 t (3 t^{2} - 7 t + 2) = 0

t = 0, t = 2 m π + 0

where m is any integer
Else

3 t^{2} - 7 t + 2 = 0

Finally, if t is

+ \infty, \frac{x}{2} = r π + \frac{π}{2}, x = 2 r π + π = (2 r + 1) π

where r is any integer
if t is

- \infty, \frac{x}{2} = s π - \frac{π}{2}, x = 2 s π - π = (2 s - 1) π

where s is any integer
In the last two case x is an odd multiple of

π

which does not satisfy the given equation

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