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

Question

asked Dec 3, 2023 157k views

1 Answer

Ramki Anba · Answer 1 · 2023-12-07T03:30:20+0000

Given:

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

Expanding using double angle formula,

$1 - 2 \sin^(2) (x) = 1 - \sin(x)$

Rearranging into a single equation,

$1 - 2 \sin^(2)(x) - 1 + \sin(x) = 0$

Combining like terms,

$- 2 \sin^(2) (x) + \sin(x) = 0$

Now multiplying each term by -1,

$2 \sin^(2) (x) - \sin(x) = 0$

Factor greatest common factors out,

$\sin(x)(2 \sin(x) - 1) = 0$

Applying Zero property rule,

$\sin(x) = 0 \: or \: 2 \sin(x) - 1 = 0$

• Solve the trignometric equation to find a particular solution,

$x = 0 \: or \: x = π$

$= > x = 2nπ \: or \: x = π + 2nπ$

• Find the union of the solution sets,

x = nπ }

• Rearrange unknown terms to the left side,

$2 \sin(x) = 1$

• Divide both sides of the equation by the coefficient of variable,

$\sin(x) = (1)/(2)$

• Solve the trignometric equation to find a particular solution,

$x = (π)/(6) \: or \: x = (5π)/(6)$

$= > x = (π)/(6) + 2nπ \: or \: x = (5π)/(6) + 2nπ,n∈Z$

}

☆ Now finding the union of both sets

$x = nπ \: or \: x = (π)/(6) + 2nπ \: or \: x = (5π)/(6) + 2nπ,n∈Z$

Hence, the answer is
$x = nπ \: or \: x = (π)/(6) + 2nπ \: or \: x = (5π)/(6) + 2nπ,n∈Z$