What is the general solution for 1-2cos^2(8x)=sin(4x) ?

The general solution for 1-2cos^2(8x)=sin(4x) is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1-2cos^2(8x)=sin(4x)

1 - 2 cos^{2} (8 x) = sin (4 x)

x = \frac{3 π + 4 πn}{8}, x = \frac{π + 12 πn}{24}, x = \frac{5 π + 12 πn}{24}, x = \frac{0.94247 \dots + 2 πn}{4}, x = \frac{π - 0.94247 \dots + 2 πn}{4}, x = \frac{- 0.31415 \dots + 2 πn}{4}, x = \frac{π + 0.31415 \dots + 2 πn}{4}

Solution steps

1 - 2 cos^{2} (8 x) = sin (4 x)

Subtract

sin (4 x)

from both sides

1 - 2 cos^{2} (8 x) - sin (4 x) = 0

Let:

u = 4 x

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

Rewrite using trig identities

- 1 - sin (u) + 8 sin^{2} (u) - 8 sin^{4} (u) = 0

Solve by substitution

sin (u) = - 1, sin (u) = \frac{1}{2}, sin (u) = \frac{1 + 5}{4}, sin (u) = \frac{1 - 5}{4}

sin (u) = - 1 : u = \frac{3 π}{2} + 2 πn

sin (u) = \frac{1}{2} : u = \frac{π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn

sin (u) = \frac{1 + 5}{4} : u = arcsin (\frac{1 + 5}{4}) + 2 πn, u = π - arcsin (\frac{1 + 5}{4}) + 2 πn

sin (u) = \frac{1 - 5}{4} : u = arcsin (\frac{1 - 5}{4}) + 2 πn, u = π + arcsin (- \frac{1 - 5}{4}) + 2 πn

Combine all the solutions

u = \frac{3 π}{2} + 2 πn, u = \frac{π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn, u = arcsin (\frac{1 + 5}{4}) + 2 πn, u = π - arcsin (\frac{1 + 5}{4}) + 2 πn, u = arcsin (\frac{1 - 5}{4}) + 2 πn, u = π + arcsin (- \frac{1 - 5}{4}) + 2 πn

Substitute back

u = 4 x

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

4 x = \frac{π}{6} + 2 πn : x = \frac{π + 12 πn}{24}

4 x = \frac{5 π}{6} + 2 πn : x = \frac{5 π + 12 πn}{24}

4 x = arcsin (\frac{1 + 5}{4}) + 2 πn : x = \frac{arcsin ( \frac{1 + 5}{4} ) + 2 πn}{4}

4 x = π - arcsin (\frac{1 + 5}{4}) + 2 πn : x = \frac{π - arcsin ( \frac{1 + 5}{4} ) + 2 πn}{4}

4 x = arcsin (\frac{1 - 5}{4}) + 2 πn : x = \frac{arcsin ( \frac{1 - 5}{4} ) + 2 πn}{4}

4 x = π + arcsin (- \frac{1 - 5}{4}) + 2 πn : x = \frac{π + arcsin ( \frac{5 - 1}{4} ) + 2 πn}{4}

x = \frac{3 π + 4 πn}{8}, x = \frac{π + 12 πn}{24}, x = \frac{5 π + 12 πn}{24}, x = \frac{arcsin ( \frac{1 + 5}{4} ) + 2 πn}{4}, x = \frac{π - arcsin ( \frac{1 + 5}{4} ) + 2 πn}{4}, x = \frac{arcsin ( \frac{1 - 5}{4} ) + 2 πn}{4}, x = \frac{π + arcsin ( \frac{5 - 1}{4} ) + 2 πn}{4}

Show solutions in decimal form

x = \frac{3 π + 4 πn}{8}, x = \frac{π + 12 πn}{24}, x = \frac{5 π + 12 πn}{24}, x = \frac{0.94247 \dots + 2 πn}{4}, x = \frac{π - 0.94247 \dots + 2 πn}{4}, x = \frac{- 0.31415 \dots + 2 πn}{4}, x = \frac{π + 0.31415 \dots + 2 πn}{4}

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

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

tan (x - 1 0^{\circ}) = 0

cos^{2} (x) = 3 sin (x) cos (x)

cos (θ) - 3 cos (2 θ) = 0

What is the general solution for 1-2cos^2(8x)=sin(4x) ?
The general solution for 1-2cos^2(8x)=sin(4x) is