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

The general solution for 2sin(x)cos(x)=cos^2(x) is x= pi/2+2pin,x=(3pi)/2+2pin,x=0.46364…+pin

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Popular >

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

Solution

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

Solution

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

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n, x = 26.56505 \dots^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

cos^{2} (x)

from both sides

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

Factor

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

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

Apply exponent rule:

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

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

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

Factor out common term

cos (x)

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

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

Solving each part separately

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

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

General solutions for

cos (x) = 0

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

2 sin (x) - cos (x) = 0 : x = arctan (\frac{1}{2}) + πn

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

Rewrite using trig identities

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

Divide both sides by

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

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

Simplify

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

Use the basic trigonometric identity:

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

2 tan (x) - 1 = 0

2 tan (x) - 1 = 0

Move

1

to the right side

2 tan (x) - 1 = 0

Add

1

to both sides

2 tan (x) - 1 + 1 = 0 + 1

Simplify

2 tan (x) = 1

2 tan (x) = 1

Divide both sides by

2

2 tan (x) = 1

Divide both sides by

2

\frac{2 tan ( x )}{2} = \frac{1}{2}

Simplify

tan (x) = \frac{1}{2}

tan (x) = \frac{1}{2}

Apply trig inverse properties

tan (x) = \frac{1}{2}

General solutions for

tan (x) = \frac{1}{2}

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

x = arctan (\frac{1}{2}) + πn

x = arctan (\frac{1}{2}) + πn

Combine all the solutions

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

Show solutions in decimal form

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

Popular Examples

-1+sin^2(a)+cos^2(a)=cos^2(a)

- 1 + sin^{2} (a) + cos^{2} (a) = cos^{2} (a)

(sin(x))/(1+cos(x))=3cot(x/2)

\frac{sin ( x )}{1 + cos ( x )} = 3 cot (\frac{x}{2})

sin(θ)=(sqrt(3))/2 ,0<= θ<= 4pi

sin (θ) = \frac{3}{2}, 0 \leq θ \leq 4 π

sin(2x)=(sqrt(3))/2 ,0<= x<= 360

sin (2 x) = \frac{3}{2}, 0^{\circ} \leq x \leq 36 0^{\circ}

3sin^2(x)+8cos^2(x)=6

3 sin^{2} (x) + 8 cos^{2} (x) = 6

Frequently Asked Questions (FAQ)

What is the general solution for 2sin(x)cos(x)=cos^2(x) ?
The general solution for 2sin(x)cos(x)=cos^2(x) is x= pi/2+2pin,x=(3pi)/2+2pin,x=0.46364…+pin