What is the general solution for (r-c)=asqrt(2)sqrt(sin(θ))+d ?

The general solution for (r-c)=asqrt(2)sqrt(sin(θ))+d is

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(r-c)=asqrt(2)sqrt(sin(θ))+d

(r - c) = a 2 sin (θ) + d

θ = arcsin (- \frac{rc}{a ^{2}} - \frac{r d}{a ^{2}} + \frac{r ^{2}}{2 a ^{2}} + \frac{c ^{2}}{2 a ^{2}} + \frac{c d}{a ^{2}} + \frac{d ^{2}}{2 a ^{2}}) + 2 πn, θ = π + arcsin (\frac{rc}{a ^{2}} + \frac{r d}{a ^{2}} - \frac{r ^{2}}{2 a ^{2}} - \frac{c ^{2}}{2 a ^{2}} - \frac{c d}{a ^{2}} - \frac{d ^{2}}{2 a ^{2}}) + 2 πn

Solution steps

(r - c) = a 2 sin (θ) + d

Switch sides

a 2 sin (θ) + d = r - c

Move

d

to the right side

a 2 sin (θ) = r - c - d

Divide both sides by

a 2

sin (θ) = \frac{2 ( r - c - d )}{2 a}

Square both sides:

sin (θ) = - \frac{rc}{a ^{2}} - \frac{r d}{a ^{2}} + \frac{r ^{2}}{2 a ^{2}} + \frac{c ^{2}}{2 a ^{2}} + \frac{c d}{a ^{2}} + \frac{d ^{2}}{2 a ^{2}}

sin (θ) = - \frac{rc}{a ^{2}} - \frac{r d}{a ^{2}} + \frac{r ^{2}}{2 a ^{2}} + \frac{c ^{2}}{2 a ^{2}} + \frac{c d}{a ^{2}} + \frac{d ^{2}}{2 a ^{2}}

Apply trig inverse properties

θ = arcsin (- \frac{rc}{a ^{2}} - \frac{r d}{a ^{2}} + \frac{r ^{2}}{2 a ^{2}} + \frac{c ^{2}}{2 a ^{2}} + \frac{c d}{a ^{2}} + \frac{d ^{2}}{2 a ^{2}}) + 2 πn, θ = π + arcsin (\frac{rc}{a ^{2}} + \frac{r d}{a ^{2}} - \frac{r ^{2}}{2 a ^{2}} - \frac{c ^{2}}{2 a ^{2}} - \frac{c d}{a ^{2}} - \frac{d ^{2}}{2 a ^{2}}) + 2 πn

8^{2} = 1 7^{2} + 1 5^{2} - 2 (17) (15) cos (A)

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

5 sin (θ) + 3 = 3

sin (3 x) = 0, 0 < x \leq 2

tan (x) = - 0.35

What is the general solution for (r-c)=asqrt(2)sqrt(sin(θ))+d ?
The general solution for (r-c)=asqrt(2)sqrt(sin(θ))+d is