"Solve tan⁡x=sec⁡42∘+3" was answered by our experts

Jamya Elliott

Answered question

2022-01-24

Solve

\tan x = {\sec 42}^{\circ} + \sqrt{3}

tsjutten20

Beginner2022-01-25Added 13 answers

{\cos 42}^{\circ} = \cos (60^{\circ} - 18^{\circ})

= {\cos 60}^{\circ} {\cos 18}^{\circ} + {\sin 60}^{\circ} {\sin 18}^{\circ}

= \frac{12}{{\cos 18}^{\circ} + \sqrt{3} {\sin 18}^{\circ}}

{\sec 42}^{\circ} = (\frac{2}{{\cos 18}^{\circ} + \sqrt{3} {\sin 18}^{\circ}}) (\frac{{\cos 18}^{\circ} - \sqrt{3} {\sin 18}^{\circ}}{{\cos 18}^{\circ} - \sqrt{3} {\sin 18}^{\circ}})

= \frac{2 ({\cos 18}^{\circ} - \sqrt{3} {\sin 18}^{\circ})}{\cos^{2} 18^{\circ} - 3 \sin^{2} 18^{\circ}}

= (\frac{2 {\sin 18}^{\circ}}{1 - 4 \sin^{2} 18^{\circ}}) ({\cot 18}^{\circ} - \sqrt{3})

Let

s = \sin (18^{\circ})

, using multiple angles formula

\sin (90^{\circ}) = \sin (5 \times 18^{\circ}) = 16 s^{5} - 20 s^{3} + 5 s = 1

16 s^{5} - 20 s^{3} + 5 s - 1 = 0

(s - 1) {(4 s^{2} + 2 s - 1)}^{2} = 0

Since

s \neq 1

, we get

4 s^{2} + 2 s - 1 = 0 \to \frac{2 s}{1 - 4 s^{2}} = 1

\tan (x) = \sec (42^{\circ}) + \sqrt{3} = (\cot (18^{\circ}) - \sqrt{3}) + \sqrt{3} = \tan (72^{\circ})

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?