Question

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

Answer 1

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,

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

{ For sin(x) = 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,

$x = nπ$}

{ For 2 sin(x) - 1 = 0 ,

• Rearrange unknown terms to the left side,

$2 \sin(x) = 1$

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

$\sin(x) = \frac{1}{2}$

• Solve the trignometric equation to find a particular solution,

$x = \frac{π}{6} \: or \: x = \frac{5π}{6}$

$= > x = \frac{π}{6} + 2nπ \: or \: x = \frac{5π}{6} + 2nπ,n∈Z$

}

☆ Now finding the union of both sets

$x = nπ \: or \: x = \frac{π}{6} + 2nπ \: or \: x = \frac{5π}{6} + 2nπ,n∈Z$

Hence, the answer is $x = nπ \: or \: x = \frac{π}{6} + 2nπ \: or \: x = \frac{5π}{6} + 2nπ,n∈Z$