Question #abd69

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Question #abd69

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Answer 1 · 2016-06-21T03:10:43

$(72^{\circ} 26 + k 180)$ $(72^@26 + k180)$
$(220^{\circ} + k 180)$ $(220^@ + k180)$

Explanation:

Apply these identities, calling $tan (\frac{x}{2}) = t$ $tan (x/2) = t$
$sin x = \frac{2 t}{1 + t^{2}}$ $sin x = (2t)/(1 + t^2)$
$cos x = \frac{1 - t^{2}}{1 + t^{2}}$ $cos x = (1 - t^2)/(1 + t^2)$
2sin x - 3cos x = 1
Replace sin x and cos x by the above identities, we get:
$4 t - 3 + 3 t^{2} = 1 + t^{2}$ $4t - 3 + 3t^2 = 1 + t^2$
$2 t^{2} + 4 t - 4 = 0$ $2t^2 + 4t - 4 = 0$
$t^{2} + 2 t - 2 = 0$ $t^2 + 2t - 2 = 0$ .
Solve this quadratic equation for t, using the improved quadratic formula (Google, Yahoo Search);
$D = d^{2} = b^{2} - 4 a c = 4 + 8 = 12$ $D = d^2 = b^2 - 4ac = 4 + 8 = 12$ --> $d = \pm 2 \sqrt{3}$ $d = +- 2sqrt3$
There are 2 real roots:
$t = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{2}{2} \pm \frac{2 \sqrt{3}}{2} = - 1 \pm \sqrt{3}$ $t = -b/(2a) +- d/(2a) = -2/2 +- (2sqrt3)/2 = -1 +- sqrt3$
$t 1 = - 1 - \sqrt{3} = - 2.73$ $t1 = -1 -sqrt3 = - 2.73$
$t 2 = - 1 + \sqrt{3} = 0.73$ $t2 = -1 + sqrt3 = 0.73$
a. $t 1 = tan (\frac{x}{2}) = - 2.73$ $t1 = tan (x/2) = -2.73$ --> $\frac{x}{2} = - 69^{\circ} 88$ $x/2 = - 69^@88$ and $\frac{x}{2} = - 69.88 + 180 = 110^{\circ} 12$ $x/2 = -69.88 + 180 = 110^@12$ --> x = 220.24 + k180
b. t2 = 0.73 --> $\frac{x}{2} = 36.13$ $x/2 = 36.13$ and $\frac{x}{2} = 36.13 + 180 = 216.13$ $x/2 = 36.13 + 180 = 216.13$ -->
$x = 72.26 + k 180$ $x = 72.26 + k180$

Answer 2 · 2016-06-21T03:52:36

Using identities of fractional angle

$sin x = \frac{2 tan (\frac{x}{2})}{1 + {tan}^{2} (\frac{x}{2})}$ $sinx=(2tan(x/2))/(1+tan^2(x/2))$

and
$cos x = \frac{1 - {tan}^{2} (\frac{x}{2})}{1 + {tan}^{2} (\frac{x}{2})}$ $cosx=(1-tan^2(x/2))/(1+tan^2(x/2))$

Given equation becomes

$2 sin x - 3 cos x = 1$ $2sinx-3cosx=1$

$= \frac{4 tan (\frac{x}{2})}{1 + {tan}^{2} (\frac{x}{2})} - 3 \frac{(1 - {tan}^{2} (\frac{x}{2}))}{1 + {tan}^{2} (\frac{x}{2})} = 1$ $=(4tan(x/2))/(1+tan^2(x/2))-3((1-tan^2(x/2)))/(1+tan^2(x/2))=1$

putting $tan (\frac{x}{2}) = t$ $tan(x/2)=t$ we get

$\Rightarrow \frac{4 t}{1 + t^{2}} - \frac{3 (1 - t^{2})}{1 + t^{2}} = 1$ $=>(4t)/(1+t^2)-(3(1-t^2))/(1+t^2)=1$

$\Rightarrow 4 t - 3 + 3 t^{2} = 1 + t^{2}$ $=>4t-3+3t^2=1+t^2$

$\Rightarrow 3 t^{2} - t^{2} + 4 t - 3 - 1 = 0$ $=>3t^2-t^2+4t-3-1=0$

$\Rightarrow 2 t^{2} + 4 t - 4 = 0$ $=>2t^2+4t-4=0$

$\Rightarrow t^{2} + 2 t - 2 = 0$ $=>t^2+2t-2=0$

$\Rightarrow t^{2} + 2 t + 1 = 3$ $=>t^2+2t+1=3$

$\Rightarrow {(t + 1)}^{2} = 3$ $=>(t+1)^2=3$

$\Rightarrow t + 1 = \pm \sqrt{3}$ $=>t+1=+-sqrt3$

$\Rightarrow t = \pm \sqrt{3} - 1$ $=>t=+-sqrt3-1$

when $t = \sqrt{3} - 1$ $t =sqrt3-1$

$tan (\frac{x}{2}) = \sqrt{3} - 1$ $tan(x/2)=sqrt3-1$

$:.x=2xxtan^-1(sqrt3-1)=2*36.2^@=72.4^@$

Or, $x=(180+72.4)=252.4^@$

Again when

$tan(x/2)=-sqrt3-1$

$=>x=2xxtan^-1(-sqrt3-1)$

$=>x=2*110.1^@=220.2^@$

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