What is the general solution for solvefor x,sin(2x)*cos(2x)=0.5 ?

The general solution for solvefor x,sin(2x)*cos(2x)=0.5 is x= pi/8+(pin)/2

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solvefor x,sin(2x)*cos(2x)=0.5

solve for

x, sin (2 x) \cdot cos (2 x) = 0.5

x = \frac{π}{8} + \frac{πn}{2}

Degrees

x = 22. 5^{\circ} + 9 0^{\circ} n

Solution steps

sin (2 x) cos (2 x) = 0.5

Rewrite using trig identities

sin (2 x) cos (2 x)

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 \cdot 2 x )}{2}

\frac{sin ( 2 \cdot 2 x )}{2} = 0.5

Multiply both sides by

2

\frac{sin ( 2 \cdot 2 x )}{2} = 0.5

Multiply both sides by

2

\frac{2 sin ( 2 \cdot 2 x )}{2} = 0.5 \cdot 2

Simplify

sin (2 \cdot 2 x) = 1

sin (2 \cdot 2 x) = 1

General solutions for

sin (2 \cdot 2 x) = 1

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}

2 \cdot 2 x = \frac{π}{2} + 2 πn

2 \cdot 2 x = \frac{π}{2} + 2 πn

Solve

2 \cdot 2 x = \frac{π}{2} + 2 πn : x = \frac{π}{8} + \frac{πn}{2}

2 \cdot 2 x = \frac{π}{2} + 2 πn

Divide both sides by

4

2 \cdot 2 x = \frac{π}{2} + 2 πn

Divide both sides by

4

\frac{2 \cdot 2 x}{4} = \frac{\frac{π}{2}}{4} + \frac{2 πn}{4}

Simplify

\frac{2 \cdot 2 x}{4} = \frac{\frac{π}{2}}{4} + \frac{2 πn}{4}

Simplify

\frac{2 \cdot 2 x}{4} : x

\frac{2 \cdot 2 x}{4}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4 x}{4}

Divide the numbers:

\frac{4}{4} = 1

= x

Simplify

\frac{\frac{π}{2}}{4} + \frac{2 πn}{4} : \frac{π}{8} + \frac{πn}{2}

\frac{\frac{π}{2}}{4} + \frac{2 πn}{4}

\frac{\frac{π}{2}}{4} = \frac{π}{8}

\frac{\frac{π}{2}}{4}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{π}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{π}{8}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

Cancel the common factor:

2

= \frac{πn}{2}

= \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

2 sin^{2} (x) + sin (x) - 3 = 0, 6

(1 - sin (2 x)) \cdot (1 - cos^{2} (x)) = 0

5 cos (x) - 10 sin (x) cos (x) = 0, 0 \leq x \leq \frac{π}{2}

∣ (211) ∣ ∣ (- 37 - 1) ∣ (cos (x)) = 0

u, x = 4 sinh (u)

What is the general solution for solvefor x,sin(2x)*cos(2x)=0.5 ?
The general solution for solvefor x,sin(2x)*cos(2x)=0.5 is x= pi/8+(pin)/2