What is the general solution for sin(2x)=-2cos^2(x),0<= x<= 2pi ?

Q: What is the general solution for sin(2x)=-2cos^2(x),0<= x<= 2pi ?

The general solution for sin(2x)=-2cos^2(x),0<= x<= 2pi is x= pi/2 ,x=(3pi)/2 ,x=(3pi)/4 ,x=(7pi)/4

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sin(2x)=-2cos^2(x),0<= x<= 2pi

Solution

sin (2 x) = - 2 cos^{2} (x), 0 \leq x \leq 2 π

Solution

x = \frac{π}{2}, x = \frac{3 π}{2}, x = \frac{3 π}{4}, x = \frac{7 π}{4}

Degrees

x = 9 0^{\circ}, x = 27 0^{\circ}, x = 13 5^{\circ}, x = 31 5^{\circ}

Solution steps

sin (2 x) = - 2 cos^{2} (x), 0 \leq x \leq 2 π

Subtract

- 2 cos^{2} (x)

from both sides

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

Factor

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

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

Apply exponent rule:

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

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

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

Factor out common term

2 cos (x)

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

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

Solving each part separately

cos (x) = 0 or cos (x) + sin (x) = 0

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

cos (x) = 0, 0 \leq x \leq 2 π

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

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{2}, x = \frac{3 π}{2}

cos (x) + sin (x) = 0, 0 \leq x \leq 2 π : x = \frac{3 π}{4}, x = \frac{7 π}{4}

cos (x) + sin (x) = 0, 0 \leq x \leq 2 π

Rewrite using trig identities

cos (x) + sin (x) = 0

Divide both sides by

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

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

Simplify

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

Use the basic trigonometric identity:

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

1 + tan (x) = 0

1 + tan (x) = 0

Move

1

to the right side

1 + tan (x) = 0

Subtract

1

from both sides

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

Simplify

tan (x) = - 1

tan (x) = - 1

General solutions for

tan (x) = - 1

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{3 π}{4}, x = \frac{7 π}{4}

Combine all the solutions

x = \frac{π}{2}, x = \frac{3 π}{2}, x = \frac{3 π}{4}, x = \frac{7 π}{4}

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Frequently Asked Questions (FAQ)

What is the general solution for sin(2x)=-2cos^2(x),0<= x<= 2pi ?
The general solution for sin(2x)=-2cos^2(x),0<= x<= 2pi is x= pi/2 ,x=(3pi)/2 ,x=(3pi)/4 ,x=(7pi)/4