What is the general solution for solvefor x,sin(x+r/4)+sin(x-r/3)=0 ?

The general solution for solvefor x,sin(x+r/4)+sin(x-r/3)=0 is x=2pin+r/(24),x=pi+2pin+r/(24)

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solvefor x,sin(x+r/4)+sin(x-r/3)=0

solve for

x, sin (x + \frac{r}{4}) + sin (x - \frac{r}{3}) = 0

x = 2 πn + \frac{r}{24}, x = π + 2 πn + \frac{r}{24}

Solution steps

sin (x + \frac{r}{4}) + sin (x - \frac{r}{3}) = 0

Rewrite using trig identities

2 cos (\frac{7 r}{24}) sin (\frac{24 x - r}{24}) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

sin (\frac{24 x - r}{24}) = 0

General solutions for

sin (\frac{24 x - r}{24}) = 0

\frac{24 x - r}{24} = 0 + 2 πn, \frac{24 x - r}{24} = π + 2 πn

Solve

\frac{24 x - r}{24} = 0 + 2 πn : x = 2 πn + \frac{r}{24}

Solve

\frac{24 x - r}{24} = π + 2 πn : x = π + 2 πn + \frac{r}{24}

x = 2 πn + \frac{r}{24}, x = π + 2 πn + \frac{r}{24}

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

x, sec (x) - sin (50 o) = 2

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

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

x, cos^{2} (x) + cos^{2} (y) cos^{2} (z) = 1

What is the general solution for solvefor x,sin(x+r/4)+sin(x-r/3)=0 ?
The general solution for solvefor x,sin(x+r/4)+sin(x-r/3)=0 is x=2pin+r/(24),x=pi+2pin+r/(24)