Expert Answer to "How do I solve the following: cos⁡(12x)=5sin⁡(3x)+9tan⁡2(x)+cot⁡2(x) "

High School
Geometry
Trigonometry
Trigonometric equations and identities

Tara Alvarado

Answered question

2021-12-30

How do I solve the following:

\cos (12 x) = 5 \sin (3 x) + 9 \tan 2 (x) + \cot 2 (x) f or x \in (0, 360)

Answer & Explanation

alkaholikd9

Beginner2021-12-31Added 37 answers

Not much different form Toby Maks

Cleveland Walters

Beginner2022-01-01Added 40 answers

a + b \geq 2 \sqrt{a b}

9 \tan^{2} x + 1 \cot^{2} x \geq 2 \sqrt{9 \tan^{2} x \times 1 \cot^{2} x} = 6

\cos (12 x) - 5 \sin (3 x) \geq 6

Implicit

\cos 12 x - 5 \sin 3 x \leq max {\cos 12 x} + max {- 5 \sin 3 x} \leq 1 + 5

and only possibilities :

{\begin{matrix} \cos 12 x = 1 \\ \sin 3 x = - 1 \end{matrix}

nick1337

Expert2022-01-08Added 777 answers

The minimum value of $9 \tan^{2} x + \cot^{2} x$ is $2 \sqrt{9 \tan^{2} x \cot^{2} x} = 6$ by AM-GM for all real x, as $u^{2} \geq 0, u \in R .$ Thus $\cos (12 x) - (5 \sin (3 x) + 9 \tan 2 x + \cot 2 x)$ is bounded above by $g (x) = \cos (12 x) - (5 \sin (3 x) + 6)$
Now g(x) is again bounded above by 1-5(-1)-6=0. Hence you just need to solve g(x)=0, but as $\cos (12 x) \leq 1, 5 \sin (3 x) + 6 \geq 1$ for all $x \in R$ , this implies:
$\cos (12 x) = 1, 5 \sin (3 x) + 6 = 1$

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