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Popular >

solvefor x,sin^2(y)+cos(x)y=k

Solution

solve for

x, sin^{2} (y) + cos (x) y = k

Solution

x = arccos (\frac{k - sin ^{2} ( y )}{y}) + 2 πn, x = - arccos (\frac{k - sin ^{2} ( y )}{y}) + 2 πn

Solution steps

sin^{2} (y) + cos (x) y = k

Move

sin^{2} (y)

to the right side

sin^{2} (y) + cos (x) y = k

Subtract

sin^{2} (y)

from both sides

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

Simplify

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

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

Divide both sides by

y; y \neq = 0

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

Divide both sides by

y; y \neq = 0

\frac{cos ( x ) y}{y} = \frac{k}{y} - \frac{sin ^{2} ( y )}{y}; y \neq = 0

Simplify

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

Simplify

\frac{cos ( x ) y}{y} : cos (x)

\frac{cos ( x ) y}{y}

Cancel the common factor:

y

= cos (x)

Simplify

\frac{k}{y} - \frac{sin ^{2} ( y )}{y} : \frac{k - sin ^{2} ( y )}{y}

\frac{k}{y} - \frac{sin ^{2} ( y )}{y}

Apply rule

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

= \frac{k - sin ^{2} ( y )}{y}

cos (x) = \frac{k - sin ^{2} ( y )}{y}; y \neq = 0

cos (x) = \frac{k - sin ^{2} ( y )}{y}; y \neq = 0

cos (x) = \frac{k - sin ^{2} ( y )}{y}; y \neq = 0

Apply trig inverse properties

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

General solutions for

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

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = - arccos (a) + 2 πn

x = arccos (\frac{k - sin ^{2} ( y )}{y}) + 2 πn, x = - arccos (\frac{k - sin ^{2} ( y )}{y}) + 2 πn

x = arccos (\frac{k - sin ^{2} ( y )}{y}) + 2 πn, x = - arccos (\frac{k - sin ^{2} ( y )}{y}) + 2 πn

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