What is the general solution for sin(θ)cos(θ)=(2sqrt(3))/5 ?

The general solution for sin(θ)cos(θ)=(2sqrt(3))/5 is No Solution for θ\in\mathbb{R}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin(θ)cos(θ)=(2sqrt(3))/5

sin (θ) cos (θ) = \frac{2 3}{5}

No Solution for θ \in R

Solution steps

sin (θ) cos (θ) = \frac{2 3}{5}

Rewrite using trig identities

sin (θ) cos (θ)

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

sin (x) cos (x) = \frac{sin ( 2 x )}{2}

= \frac{sin ( 2 θ )}{2}

\frac{sin ( 2 θ )}{2} = \frac{2 3}{5}

Multiply both sides by

2

\frac{sin ( 2 θ )}{2} = \frac{2 3}{5}

Multiply both sides by

2

\frac{2 sin ( 2 θ )}{2} = \frac{2 \cdot 2 3}{5}

Simplify

\frac{2 sin ( 2 θ )}{2} = \frac{2 \cdot 2 3}{5}

Simplify

\frac{2 sin ( 2 θ )}{2} : sin (2 θ)

\frac{2 sin ( 2 θ )}{2}

Divide the numbers:

\frac{2}{2} = 1

= sin (2 θ)

Simplify

\frac{2 \cdot 2 3}{5} : \frac{4 3}{5}

\frac{2 \cdot 2 3}{5}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4 3}{5}

sin (2 θ) = \frac{4 3}{5}

sin (2 θ) = \frac{4 3}{5}

sin (2 θ) = \frac{4 3}{5}

- 1 \leq sin (x) \leq 1

No Solution for θ \in R

tan (2 5^{\circ}) = cot (2 x + 5)

sin (a) = 0.4

sin (a) = 0.2

sin (2 x) = 1.47 (sin (x))^{2}

2 sin^{2} (u) = 1 +, u

What is the general solution for sin(θ)cos(θ)=(2sqrt(3))/5 ?
The general solution for sin(θ)cos(θ)=(2sqrt(3))/5 is No Solution for θ\in\mathbb{R}