Question 132

If $$x\sin^{2} 60^{\circ}-\frac{3}{2} \sec60^{\circ} \tan^{2}30^{\circ}$$ + $$\frac{4}{5}\sin^{2}45^{\circ}\tan^{2}60^{\circ}=0$$ then x is

Solution

Remember that
sin 60° = √3/2, sin 45° = 1/√2
sin 30° = 1/2
cos 30° = √3/2
tan 30° = 1/√3
sec 60° = 2

tan 60° = √3

So the above euation becomes

x(√3/2)^2-(3/2)*2*(1/√3)^2+(4/5)*1(/√2)^2*(√3)^2=0

3x/4-1+6/5=0

x=(1-6/5) *4/3

x=-4/15

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