Question-157080

Question Number 157080 by amin96 last updated on 19/Oct/21

Commented by cortano last updated on 20/Oct/21

\Rightarrow 1 + \frac{1 - \cos 2 x}{2} = \frac{3 \sin 2 x}{2}

\Rightarrow 3 - \cos 2 x = 3 \sin 2 x

\Rightarrow 3 - \cos 2 x = 3 \sqrt{1 - \cos^{2} (2 x)}

\Rightarrow 9 + \cos^{2} (2 x) - 6 \cos (2 x) = 9 - 9 \cos^{2} (2 x)

\Rightarrow 10 \cos^{2} (2 x) - 6 \cos (2 x) = 0

\Rightarrow \cos 2 x (5 \cos (2 x) - 3) = 0

{\begin{cases} \cos 2 x = 0 \Rightarrow x = \pm \frac{π}{4} + n π \\ \cos 2 x = \frac{3}{5} \Rightarrow \tan (2 x) = \frac{4}{3} \end{cases}

\Rightarrow \frac{2 \tan x}{1 - \tan^{2} x} = \frac{4}{3}

\Rightarrow 6 \tan x = 4 - 4 \tan^{2} x

\Rightarrow 2 \tan^{2} x + 3 \tan x - 2 = 0

\Rightarrow (2 \tan x - 1) (\tan x + 2) = 0

Σ \tan (x_{i}) = - \frac{3}{2}

Commented by mr W last updated on 19/Oct/21

\tan x = \frac{1}{2} o r 1

\Rightarrow Σ \tan x = \frac{3}{2}

Answered by mr W last updated on 19/Oct/21

1 + \frac{1}{2} (1 - \cos 2 x) = \frac{3}{2} \sin 2 x

3 \sin 2 x + \cos 2 x = 3

l e t t = \tan x

3 \times \frac{2 t}{1 + t^{2}} + \frac{1 - t^{2}}{1 + t^{2}} = 3

2 t^{2} - 3 t + 1 = 0

Σ t = Σ \tan x = \frac{3}{2}

Answered by amin96 last updated on 19/Oct/21

\frac{1}{\cos^{2} (x)} + \frac{\sin^{2} (x)}{\cos^{2} (x)} = 3 \frac{\sin (x) \cos (x)}{\cos^{2} (x)}

1 + {tg}^{2} (x) + t^{2} (x) = 3 tg (x)

2 {tg}^{2} (x) - 3 tg (x) + 1 = 0

2 a^{2} - 3 a + 1 = 0 {\begin{cases} a_{1} = tg {(x)}_{1} = \frac{1}{2} \\ a_{2} = tg {(x)}_{2} = 1 \end{cases}

a_{1} + a_{2} = 1, 5

I think a solution like this would be easier.

Commented by amin96 last updated on 19/Oct/21

just occurred to me)

Answered by lyubita last updated on 19/Oct/21

\sin^{2} x + \cos^{2} x + \sin^{2} x = 3 \sin x \cos x

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

(2 \sin x - \cos x) (\sin x - \cos x) = 0

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

\tan x = \frac{1}{2} \tan x = 1

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