What is the general solution for solvefor y,7y=sin^2(x)-4sin(x)+3 ?

The general solution for solvefor y,7y=sin^2(x)-4sin(x)+3 is y=(sin^2(x)-4sin(x)+3)/7

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solvefor y,7y=sin^2(x)-4sin(x)+3

solve for

y, 7 y = sin^{2} (x) - 4 sin (x) + 3

y = \frac{sin ^{2} ( x ) - 4 sin ( x ) + 3}{7}

Solution steps

7 y = sin^{2} (x) - 4 sin (x) + 3

Divide both sides by

7

7 y = sin^{2} (x) - 4 sin (x) + 3

Divide both sides by

7

\frac{7 y}{7} = \frac{sin ^{2} ( x )}{7} - \frac{4 sin ( x )}{7} + \frac{3}{7}

Simplify

\frac{7 y}{7} = \frac{sin ^{2} ( x )}{7} - \frac{4 sin ( x )}{7} + \frac{3}{7}

Simplify

\frac{7 y}{7} : y

\frac{7 y}{7}

Divide the numbers:

\frac{7}{7} = 1

= y

Simplify

\frac{sin ^{2} ( x )}{7} - \frac{4 sin ( x )}{7} + \frac{3}{7} : \frac{sin ^{2} ( x ) - 4 sin ( x ) + 3}{7}

\frac{sin ^{2} ( x )}{7} - \frac{4 sin ( x )}{7} + \frac{3}{7}

Apply rule

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

= \frac{sin ^{2} ( x ) - 4 sin ( x ) + 3}{7}

y = \frac{sin ^{2} ( x ) - 4 sin ( x ) + 3}{7}

y = \frac{sin ^{2} ( x ) - 4 sin ( x ) + 3}{7}

y = \frac{sin ^{2} ( x ) - 4 sin ( x ) + 3}{7}

tan (1 1^{\circ}) = \frac{x}{12}

c, cos (x) - sin (x) + 3 sin^{2} (x) - 1 = 0

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

a, sin (0) = \frac{a}{h}

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

What is the general solution for solvefor y,7y=sin^2(x)-4sin(x)+3 ?
The general solution for solvefor y,7y=sin^2(x)-4sin(x)+3 is y=(sin^2(x)-4sin(x)+3)/7