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Popular >

2sin(2θ)=cos(2θ),0<= θ<= 2pi

Solution

2 sin (2 θ) = cos (2 θ), 0 \leq θ \leq 2 π

Solution

θ = \frac{0.46364 \dots}{2}, θ = \frac{0.46364 \dots + π}{2}, θ = \frac{0.46364 \dots + 2 π}{2}, θ = \frac{0.46364 \dots + 3 π}{2}

+1

Degrees

θ = 13.28252 \dots^{\circ}, θ = 103.28252 \dots^{\circ}, θ = 193.28252 \dots^{\circ}, θ = 283.28252 \dots^{\circ}

Solution steps

2 sin (2 θ) = cos (2 θ), 0 \leq θ \leq 2 π

Rewrite using trig identities

2 sin (2 θ) = cos (2 θ)

Divide both sides by

cos (2 θ), cos (2 θ) \neq = 0

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

Simplify

\frac{2 sin ( 2 θ )}{cos ( 2 θ )} = 1

Use the basic trigonometric identity:

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

2 tan (2 θ) = 1

2 tan (2 θ) = 1

Divide both sides by

2

2 tan (2 θ) = 1

Divide both sides by

2

\frac{2 tan ( 2 θ )}{2} = \frac{1}{2}

Simplify

tan (2 θ) = \frac{1}{2}

tan (2 θ) = \frac{1}{2}

Apply trig inverse properties

tan (2 θ) = \frac{1}{2}

General solutions for

tan (2 θ) = \frac{1}{2}

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

2 θ = arctan (\frac{1}{2}) + πn

2 θ = arctan (\frac{1}{2}) + πn

Solve

2 θ = arctan (\frac{1}{2}) + πn : θ = \frac{arctan ( \frac{1}{2} )}{2} + \frac{πn}{2}

2 θ = arctan (\frac{1}{2}) + πn

Divide both sides by

2

2 θ = arctan (\frac{1}{2}) + πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{arctan ( \frac{1}{2} )}{2} + \frac{πn}{2}

Simplify

θ = \frac{arctan ( \frac{1}{2} )}{2} + \frac{πn}{2}

θ = \frac{arctan ( \frac{1}{2} )}{2} + \frac{πn}{2}

θ = \frac{arctan ( \frac{1}{2} )}{2} + \frac{πn}{2}

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{arctan ( \frac{1}{2} )}{2}, θ = \frac{arctan ( \frac{1}{2} ) + π}{2}, θ = \frac{arctan ( \frac{1}{2} ) + 2 π}{2}, θ = \frac{arctan ( \frac{1}{2} ) + 3 π}{2}

Show solutions in decimal form

θ = \frac{0.46364 \dots}{2}, θ = \frac{0.46364 \dots + π}{2}, θ = \frac{0.46364 \dots + 2 π}{2}, θ = \frac{0.46364 \dots + 3 π}{2}

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