Solve this trigonometric equation sin⁡2x−3cos⁡2x=2 I tried dividing both sides with cos⁡2x but then I win 2cos⁡2x

12sin⁡2x−32cos⁡2x=1 sin⁡2xcos⁡π3−cos⁡2xsin⁡π3=1 sin⁡(2x−π3)=1 I hope you can solve further.

MySolution:: Given sin⁡2x−3cos⁡2x=2 We can write it as sin⁡2x⋅12−cos⁡2x⋅32=1⇒sin⁡(2x−π3)=1=sin⁡π2 Above we have used the formula sin⁡α⋅cos⁡β−cos⁡α⋅sin⁡β=sin⁡(α−β) So 2x−π3=nπ+(−1)n⋅π2⇒x=nπ2+(−1)n⋅π4+π6 Where n∈Z

Divide by 2 to get 1=12sin⁡(2x)−32cos⁡(2x) then find an angle y for which cos⁡y=12 and sin⁡y=32 and remember that sin⁡(a−b)=sin⁡acos⁡b−cos⁡asin⁡b

12sin⁡2x−32cos⁡2x=1 sin⁡2xcos⁡π3−cos⁡2xsin⁡π3=1 sin⁡(2x−π3)=1 I hope you can solve further.

MySolution:: Given sin⁡2x−3cos⁡2x=2 We can write it as sin⁡2x⋅12−cos⁡2x⋅32=1⇒sin⁡(2x−π3)=1=sin⁡π2 Above we have used the formula sin⁡α⋅cos⁡β−cos⁡α⋅sin⁡β=sin⁡(α−β) So 2x−π3=nπ+(−1)n⋅π2⇒x=nπ2+(−1)n⋅π4+π6 Where n∈Z

Divide by 2 to get 1=12sin⁡(2x)−32cos⁡(2x) then find an angle y for which cos⁡y=12 and sin⁡y=32 and remember that sin⁡(a−b)=sin⁡acos⁡b−cos⁡asin⁡b

Expert Answer to "Solve this trigonometric equation sin⁡2x−3cos⁡2x=2 I tried dividing..."

High School
Geometry
Trigonometry
Trigonometric equations and identities

Dowqueuestbew1j

Answered question

2021-12-27

Solve this trigonometric equation

\sin 2 x - \sqrt{3} \cos 2 x = 2

I tried dividing both sides with

\cos 2 x

but then I win

\frac{2}{\cos 2 x}

Answer & Explanation

sukljama2

Beginner2021-12-28Added 32 answers

\frac{12}{\sin 2} x - \frac{\sqrt{3}}{2} \cos 2 x = 1

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

\sin (2 x - \frac{π}{3}) = 1

I hope you can solve further.

abonirali59

Beginner2021-12-29Added 35 answers

MySolution:: Given

\sin 2 x - \sqrt{3} \cos 2 x = 2

We can write it as

\sin 2 x \cdot \frac{12}{-} \cos 2 x \cdot \frac{\sqrt{3}}{2} = 1 \Rightarrow \sin (2 x - \frac{π}{3}) = 1 = \sin \frac{π}{2}

Above we have used the formula

\sin α \cdot \cos β - \cos α \cdot \sin β = \sin (α - β)

2 x - \frac{π}{3} = n π + {(- 1)}^{n} \cdot \frac{π}{2} \Rightarrow x = \frac{n π}{2} + {(- 1)}^{n} \cdot \frac{π}{4} + \frac{π}{6}

Where

n \in Z

nick1337

Expert2022-01-08Added 777 answers

Divide by 2 to get
$1 = \frac{1}{2} \sin (2 x) - \frac{\sqrt{3}}{2} \cos (2 x)$
then find an angle y for which $\cos y = \frac{1}{2}$ and $\sin y = \frac{\sqrt{3}}{2}$ and remember that
$\sin (a - b) = \sin a \cos b - \cos a \sin b$

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