Question #5bc91 Trigonometry 1 Answer Cem Sentin Dec 1, 2017 x_1=pi/12, x_2=(3pi)/4 and x_3=(17pi)/12 Explanation: 2(cos3x)^2-3sqrt2cos3x+2=0 (sqrt2cos3x)^2-3sqrt2cos3x+2=0 After using y=sqrt2cos2x transformation, this equation became, y^2-3y+2=0 or (y-1)*(y-2)=0 Hence y_1=1 and y_2=2 For y=2, sqrt2*cos3x=2 or cos3x=sqrt2. This is impossible. For y=1, sqrt2*cos3x=1 or cos3x=sqrt2/2. Hence 3x=pi/4+2pi*k or x=pi/12+(2pi*k)/3 Thus solutions of it are x_1=pi/12 (for k=0), x_2=(3pi)/4 (for k=1) and x_3=(17pi)/12 (for k=2) Answer link Related questions How do I determine the molecular shape of a molecule? What is the lewis structure for co2? What is the lewis structure for hcn? How is vsepr used to classify molecules? What are the units used for the ideal gas law? How does Charle's law relate to breathing? What is the ideal gas law constant? How do you calculate the ideal gas law constant? How do you find density in the ideal gas law? Does ideal gas law apply to liquids? Impact of this question 818 views around the world You can reuse this answer Creative Commons License