Trigonometry

[Mary Smith]

Can't understad this...3 sin²o +5 cos o - 1 = 0 help please?

[Guest]

Can we write the equation as follows : 3[tex]sin^{2}[/tex][tex]\alpha[/tex]+5cos[tex]\alpha[/tex]-1=0
Is there an answer ?

[Guest]

[tex]\sin^2 \alpha = \cos^2 \alpha[/tex]
Substitute [tex]cos \alpha = x[/tex]
and solve as a quadratic equation
Check if the both roots are between -1 and 1.
Only one of them should be in that interval and then [tex]cos \alpha =[/tex] some numeric value.
When we know what is tha value of cos we can check in a trigonometrc table what is the value of alpha.

[Guest]

Could you write the solution, please?

[Guest]

I am the first guest.
The second guest was wrong, because [tex]sin^{2}[/tex][tex]\alpha[/tex]=1-[tex]cos^{2}[/tex][tex]\alpha[/tex]

I've already decided the equation and got: cos[tex]\alpha[/tex]=-[tex]\frac{1}{3}[/tex]
Then [tex]\alpha[/tex]=

[Guest]

Every student that study trigonometry must have a reference book with values of sin, cos, tan and cot for the angles between 0 and 90 degrees.
I see that [tex]\cos 70^\circ 30' = 0.3338[/tex].
You can also use a calculator.

Have in mind that if we know [tex]cos \alpha[/tex] we know [tex]\alpha[/tex]

[tex]\cos(180-\alpha) = -\cos(\alpha)[/tex]
[tex]\alpha = 180 - 70.30 = 109.70[/tex]

Is it clear?

[Guest]

3[tex]sin^{2}[/tex][tex]\alpha[/tex]+5cos[tex]\alpha[/tex]-1=0
3(1-[tex]cos^{2}[/tex][tex]\alpha[/tex])+5cos[tex]\alpha[/tex]-1=0 /.(-1)[tex]\ne[/tex]0

3[tex]cos^{2}[/tex][tex]\alpha[/tex]-5cos[tex]\alpha[/tex]-2=0 ; we apply cos[tex]\alpha[/tex]=x [tex]\Rightarrow[/tex] 3[tex]x^{2}[/tex]-5x-2=0

D=25+24=49 ; [tex]x_{1,2 }[/tex]=[tex]\frac{5\pm\sqrt{49}}{2.3}[/tex] [tex]\Rightarrow[/tex] [tex]x_{1 }[/tex]=2 ,[tex]x_{2 }[/tex]=-[tex]\frac{1}{3}[/tex]
cos[tex]\alpha[/tex]=2[tex]\notin[/tex][-1;1] falls out ; cos[tex]\alpha[/tex]=-[tex]\frac{1}{3}[/tex][tex]\in[/tex][-1;1]
-cos[tex]\alpha[/tex]=0,333...
cos(180[tex]^\circ[/tex]-[tex]\alpha[/tex])=cos 70[tex]^\circ[/tex] 30'
180[tex]^\circ[/tex]-[tex]\alpha[/tex]=70[tex]^\circ[/tex]30'
109[tex]^\circ[/tex]30'=[tex]\alpha[/tex] [I think 30'+30'=1[tex]^\circ[/tex] ]
My answer: 109[tex]^\circ[/tex]30'