How do you solve $x = h + a \cdot cos (t) \cdot cos (s) - b \cdot sin (t) \cdot sin (s)$ $x=h+acos(t)cos(s)-bsin(t)sin(s)$ for $t$ $t$ ?

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Answer 1 · 2018-08-09T01:09:16

See explicit t = g ( x, s )#, in the explanation.

Using c = cos s and d = sin s,

${(x - h)}^{2} = c^{2} a^{2} (1 - d^{2}) + (1 - c^{2}) b^{2} d^{2}$ $( x - h )^2 = c^2a^2( 1 - d^2 ) + ( 1 - c^2 ) b^2 d^2$

$- 2 c d a b \sqrt{(1 - c^{2}) (1 - d^{2})}$ $- 2cdab sqrt(( 1 - c^2 )( 1 - d^2))$ . Reorganizing and squaring,

${({(x - h)}^{2} - c^{2} a^{2} (1 - d^{2}) - (1 - c^{2}) b^{2} d^{2})}^{2}$ $(( x - h )^2 - c^2a^2( 1 - d^2 ) - ( 1 - c^2 ) b^2 d^2)^2$

$= 4 {(d a b)}^{2} (1 - d^{2}) c^{2} ((1 - c^{2})$ $= 4(dab)^2(1-d^2) c^2 ((1-c^2)$

Befitting ( $| c | \leq 1$ $abs c <= 1$ ) solutions ${cos α}$ ${ cos alpha }$ of this

biquadratic in c, lead to

$c = cos t = cos α$ $c = cos t = cos alpha$

$\Rightarrow$ $rArr$ , piecewise,

$t = 2 k π \pm α$ $t = 2kpi +-alpha$ rad.

Example: a = b = 1 and h = 0.

$x = cos t cos s - sin t sin s$ $x = cos t cos s - sin t sin s$ and

(x^2 -c^2 - d^2 + 2c^2 d^2 )^2#

$= 4 c^{2} (1 - c^{2}) d^{2} (1 - d^{2})$ $= 4 c^2 (1-c^2) d^2 (1-d^2)$

Solve this quadratic in $c^{2}$ $c^2$ for c = cos t.

I think, I have paved the way.