What is the general solution for 4sin^2(x)-cos^2(x)=0 ?

The general solution for 4sin^2(x)-cos^2(x)=0 is x=-0.46364…+pin,x=0.46364…+pin

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Popular >

4sin^2(x)-cos^2(x)=0

Solution

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

Solution

x = - 0.46364 \dots + πn, x = 0.46364 \dots + πn

Degrees

x = - 26.56505 \dots^{\circ} + 18 0^{\circ} n, x = 26.56505 \dots^{\circ} + 18 0^{\circ} n

Solution steps

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

Factor

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

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

Rewrite

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

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

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

Rewrite

4

2^{2}

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

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

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

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

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

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

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

Solving each part separately

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

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

Subtract

1

from 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

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

Show solutions in decimal form

x = - 0.46364 \dots + πn, x = 0.46364 \dots + πn

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sin (7 x) = cos (2 x)

Frequently Asked Questions (FAQ)

What is the general solution for 4sin^2(x)-cos^2(x)=0 ?
The general solution for 4sin^2(x)-cos^2(x)=0 is x=-0.46364…+pin,x=0.46364…+pin