What is the general solution for sin(2x)=1.47(sin(x))^2 ?

The general solution for sin(2x)=1.47(sin(x))^2 is x=2pin,x=pi+2pin,x=0.93696…+pin

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

Solution

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

Solution

x = 2 πn, x = π + 2 πn, x = 0.93696 \dots + πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 53.68411 \dots^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

1.47 (sin (x))^{2}

from both sides

sin (2 x) - 1.47 sin^{2} (x) = 0

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

- 1.47 sin^{2} (x) + 2 cos (x) sin (x) = 0

Factor

- 1.47 sin^{2} (x) + 2 cos (x) sin (x) : sin (x) (- 1.47 sin (x) + 2 cos (x))

- 1.47 sin^{2} (x) + 2 cos (x) sin (x)

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

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

= - 1.47 sin (x) sin (x) + 2 sin (x) cos (x)

Factor out common term

sin (x)

= sin (x) (- 1.47 sin (x) + 2 cos (x))

sin (x) (- 1.47 sin (x) + 2 cos (x)) = 0

Solving each part separately

sin (x) = 0 or - 1.47 sin (x) + 2 cos (x) = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

- 1.47 sin (x) + 2 cos (x) = 0 : x = arctan (\frac{200}{147}) + πn

- 1.47 sin (x) + 2 cos (x) = 0

Rewrite using trig identities

- 1.47 sin (x) + 2 cos (x) = 0

Divide both sides by

cos (x), cos (x) \neq = 0

\frac{- 1.47 sin ( x ) + 2 cos ( x )}{cos ( x )} = \frac{0}{cos ( x )}

Simplify

- \frac{1.47 sin ( x )}{cos ( x )} + 2 = 0

Use the basic trigonometric identity:

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

- 1.47 tan (x) + 2 = 0

- 1.47 tan (x) + 2 = 0

Multiply both sides by

100

- 1.47 tan (x) + 2 = 0

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere are

2

digits to the right of the decimal point, therefore multiply by

100

- 1.47 tan (x) \cdot 100 + 2 \cdot 100 = 0 \cdot 100

Refine

- 147 tan (x) + 200 = 0

- 147 tan (x) + 200 = 0

Move

200

to the right side

- 147 tan (x) + 200 = 0

Subtract

200

from both sides

- 147 tan (x) + 200 - 200 = 0 - 200

Simplify

- 147 tan (x) = - 200

- 147 tan (x) = - 200

Divide both sides by

- 147

- 147 tan (x) = - 200

Divide both sides by

- 147

\frac{- 147 tan ( x )}{- 147} = \frac{- 200}{- 147}

Simplify

tan (x) = \frac{200}{147}

tan (x) = \frac{200}{147}

Apply trig inverse properties

tan (x) = \frac{200}{147}

General solutions for

tan (x) = \frac{200}{147}

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan (\frac{200}{147}) + πn

x = arctan (\frac{200}{147}) + πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = arctan (\frac{200}{147}) + πn

Show solutions in decimal form

x = 2 πn, x = π + 2 πn, x = 0.93696 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for sin(2x)=1.47(sin(x))^2 ?
The general solution for sin(2x)=1.47(sin(x))^2 is x=2pin,x=pi+2pin,x=0.93696…+pin