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

The general solution for cos^2(x)=(1-sin(2x))/2 is x=(3pi)/8+(pin)/2

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

Solution

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

Solution

x = \frac{3 π}{8} + \frac{πn}{2}

Degrees

x = 67. 5^{\circ} + 9 0^{\circ} n

Solution steps

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

Subtract

\frac{1 - sin ( 2 x )}{2}

from both sides

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Expand

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

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

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

- (1 - sin (2 x)) : - 1 + sin (2 x)

- (1 - sin (2 x))

Distribute parentheses

= - (1) - (- sin (2 x))

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - 1 + sin (2 x)

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

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

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

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

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

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{3 π}{4} + πn

2 x = \frac{3 π}{4} + πn

Solve

2 x = \frac{3 π}{4} + πn : x = \frac{3 π}{8} + \frac{πn}{2}

2 x = \frac{3 π}{4} + πn

Divide both sides by

2

2 x = \frac{3 π}{4} + πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{3 π}{4}}{2} + \frac{πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{3 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{3 π}{4}}{2} + \frac{πn}{2} : \frac{3 π}{8} + \frac{πn}{2}

\frac{\frac{3 π}{4}}{2} + \frac{πn}{2}

\frac{\frac{3 π}{4}}{2} = \frac{3 π}{8}

\frac{\frac{3 π}{4}}{2}

Apply the fraction rule:

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

= \frac{3 π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{3 π}{8}

= \frac{3 π}{8} + \frac{πn}{2}

x = \frac{3 π}{8} + \frac{πn}{2}

x = \frac{3 π}{8} + \frac{πn}{2}

x = \frac{3 π}{8} + \frac{πn}{2}

x = \frac{3 π}{8} + \frac{πn}{2}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for cos^2(x)=(1-sin(2x))/2 ?
The general solution for cos^2(x)=(1-sin(2x))/2 is x=(3pi)/8+(pin)/2