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I am having some problems with the two following questions. I was wondering if someone could check my work or offer some insight.

Solve the equation $22\cos^4{(\theta/2)}=3$ for all positive values of ${\theta} $ between 0 and 360 degrees

My solution is:

$$\cos^4({\theta}/2)=(3/22) \\ \cos({\theta}/2)=({3/22})^{1/4} \\ -((1+ \cos{\theta)}/2)^{1/2}=({3/22})^{1/4} \\ \cos{\theta}=-(((3/22)^{1/4*2}*2)-1)\\ {\theta}=74.8$$

Values for ${\theta}=105, 195$

Solve the equation $\sin{\theta}=1-6 \cos{\theta}$ for all positive values of $\theta$ between 0 and 360 degrees.

My solution is:

$${\sin{\theta}}^{2}=({1-6 \cos{\theta}})^{2} \\ {\sin^2}{\theta}=1-12\cos{\theta}+36\cos^2{\theta} \\ 1-\cos^2{\theta}=1-12\cos{\theta}+36\cos^2{\theta} \\ 0=37 \cos^2{\theta}-12 \cos{\theta} \\ 0= \cos{\theta}(37 \cos{\theta}-12) \\ \cos{\theta}=0=90, 270 ~\text{deg} \\ \cos{\theta}=(12/37) \\ \theta=71.1, 289 ~\text{deg}$$

bjcolby15
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persimonns
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1 Answers1

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$$\implies\cos^2\dfrac\theta2=\sqrt{\dfrac3{22}}$$

$$\implies\cos\theta=2\cos^2\dfrac\theta2-1=\sqrt{\dfrac6{11}}-1<0$$

$$\theta=360^\circ n\pm\arccos\left(\sqrt{\dfrac6{11}}-1\right)$$

Finally, use How do I prove that $\arccos(x) + \arccos(-x)=\pi$ when $x \in [-1,1]$?

lab bhattacharjee
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