Answered By Harsh S

Part A -

Assume # = Theta 

6 Cos# Sin# - 2 Cos# = 0;

So, 2Cos# (3Sin# - 1) = 0;

So, dividing both sides by 2cos#. we get

3Sin# - 1 = 0;

So, Sin# = 1/3;

Therefore # = 19.47*

Answered By Majid B

a)    $0<\theta<2\pi$0<θ<‍2π 

$6\left(\cos\theta\right)\left(\sin\theta\right)-2\left(\cos\theta\right)=0$6(cosθ)(sinθ)−2(‍cosθ)=0      =>      $2\left(\cos\theta\right)\left(3\left(\sin\theta\right)-1\right)=0$2(cosθ)(3(sinθ)‍−1)=0

case1:  $2\left(\cos\theta\right)=0$2(cosθ)=0     =>     $\cos\theta=0$cosθ=‍0     =>      $\theta=\frac{\pi}{2},\frac{3\pi}{2}$θ=π2 ,3π2  

case2:  $3\left(\sin\theta\right)-1=0$3(sinθ)−1=0    =>   $\sin\theta=\frac{1}{3}$sinθ=‍13     =>     $\theta=0.340\left(rad\right),2.802\left(rad\right)$θ=0.340(rad),2.802(rad) 

b)   General solution

case1: $\cos\theta=0$cosθ=0   =>    $\theta=\left(2n+1\right)\left(\frac{\pi}{2}\right)$θ=(2n+1)(π2 )  ,  $n\text{∈}I=\left\{0,\pm1,\pm2,...\right\}$n∈I={0,±1,±‍2,...}  

case2:  $\sin\theta=\frac{1}{3}$sinθ‍=13    =>   $\theta=2n\pi+0.340$θ=2n‍π+0.340    ,    $\theta=2n\pi+2.802$θ=2nπ+2.802   ,     $n\text{∈}I=\left\{0,\pm1,\pm2,...\right\}$n∈I={0,±1,±2,...}