What is the general solution for 2cos^2(x)-sin(2x)-1=0 ?

The general solution for 2cos^2(x)-sin(2x)-1=0 is x= pi/8+(pin)/2

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Popular >

2cos^2(x)-sin(2x)-1=0

Solution

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

Solution

x = \frac{π}{8} + \frac{πn}{2}

Degrees

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

Solution steps

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

Rewrite using trig identities

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

Use the Double Angle identity:

2 cos^{2} (x) - 1 = cos (2 x)

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

Divide both sides by

cos (2 x), cos (2 x) \neq = 0

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

Simplify

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

Use the basic trigonometric identity:

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

1 - tan (2 x) = 0

1 - tan (2 x) = 0

Move

1

to the right side

1 - tan (2 x) = 0

Subtract

1

from both sides

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

Simplify

- tan (2 x) = - 1

- tan (2 x) = - 1

Divide both sides by

- 1

- tan (2 x) = - 1

Divide both sides by

- 1

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

Simplify

tan (2 x) = 1

tan (2 x) = 1

General solutions for

tan (2 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}

2 x = \frac{π}{4} + πn

2 x = \frac{π}{4} + πn

Solve

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

2 x = \frac{π}{4} + πn

Divide both sides by

2

2 x = \frac{π}{4} + πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

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

\frac{\frac{π}{4}}{2} + \frac{πn}{2}

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{π}{8}

= \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}

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(x)-sin(2x)-1=0 ?
The general solution for 2cos^2(x)-sin(2x)-1=0 is x= pi/8+(pin)/2