Find the maximum angle X in the range 0∘≤x≤360∘ which satisfies the equation cos2(2x)+3sin⁡(2x)−74=0

Note you have cos22x=1−sin22x; with sin⁡2x=t you get1−t2+t3−74=0so the equation can be rewritten4t2−4t3+3=0This is actually (2t−3)2=0, so the only root is t=32. Hence 2x=π3+2kπ or 2x=2π3+2kπ. Hencex=π6+kπ or x=π3+kπIf you prefer degrees,x=30∘+k180∘ or x=60∘+k180∘It shouldn't be difficult to find the maximum solution.

"Find the maximum angle X in the range..." solved and explained

High School
Geometry
Trigonometry
Trigonometric Functions

Maximus George

Answered question

2022-01-24

Find the maximum angle X in the range

0^{\circ} \leq x \leq 360^{\circ}

which satisfies the equation

\cos^{2} (2 x) + \sqrt{3} \sin (2 x) - \frac{74}{=} 0

Answer & Explanation

basgrwthej

Beginner2022-01-25Added 13 answers

Note you have

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

; with

\sin 2 x = t

you get

1 - t^{2} + t \sqrt{3} - \frac{74}{=} 0

so the equation can be rewritten

4 t^{2} - 4 t \sqrt{3} + 3 = 0

This is actually

{(2 t - \sqrt{3})}^{2} = 0

, so the only root is

t = \frac{\sqrt{3}}{2}

. Hence

2 x = \frac{π}{3} + 2 k π or 2 x = \frac{2 π}{3} + 2 k π

. Hence

x = \frac{π}{6} + k π or x = \frac{π}{3} + k π

If you prefer degrees,

x = 30^{\circ} + k 180^{\circ} or x = 60^{\circ} + k 180^{\circ}

It shouldn't be difficult to find the maximum solution.

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