Need Help With This Trig Equation

[Scremin34Egl]

Solve for x (-180<x<180)

sinx(2cosx + sin60) = cosx*cos60

2sinxcosx + sqrt3/2*sinx = 1/2cosx........stuck

 

[Scremin34Egl]

Any Help?????

 

[DrPhil]

Solve for x (-180<x<180)

sinx(2cosx + sin60) = cosx*cos60

2sinxcosx + sqrt3/2*sinx = 1/2cosx........stuck

I'm going to wait till later to put in numeric values for sine and cosine of 60°.

2 sinx cosx = cosx cos60 - sinx sin60
sin(2x) = cos(x + 60)

Can you proceed from there?

 

[Deleted member 4993]

Another hint:

sin(Θ) = cos(π/2 - Θ) and

sin(Θ) = cos(3π/2 + Θ)

 

[Scremin34Egl]

I'm going to wait till later to put in numeric values for sine and cosine of 60°.

2 sinx cosx = cosx cos60 - sinx sin60
sin(2x) = cos(x + 60)

Can you proceed from there?

sin(2x) = sin(90-(x+60))
2x = 90-x-60 or 2x = 180-(90-x-60)
3x = 30 + k.360 or x = 150+k.360
x = 10 + k.120

Correct?

 

[DrPhil]

sin(2x) = sin(90-(x+60))
2x = 90-x-60 or 2x = 180-(90-x-60)
3x = 30 + k.360 or x = 150+k.360
x = 10 + k.120

Correct?

Correct, but you only need to report the three angles within -180°<x<180°

 

[Scremin34Egl]

Correct, but you only need to report the three angles within -180°<x<180°

so x = 150,10,130

Could you help me with this one,
1) It is given that cos(2x) = cos^2(x) - sin^2(x)

1.1) Similarly, write down a formula for cos(x)

cos(2x) = 2cos^2(x) - 1
cos^2(x) = 1/2
cos(x) = sqrt(1/2)

Dont think thats right

1.2) Hence show that cos(x) = 1-tan^2*1/2(x) / 1+tan^2*1/2(x)

 

[DrPhil]

so x = 150,10,130...10, 130, -110

Could you help me with this one,
1) It is given that cos(2x) = cos^2(x) - sin^2(x)

1.1) Similarly, write down a formula for cos(x)

cos(2x) = 2cos^2(x) - 1
cos^2(x) = 1/2
cos(x) = sqrt(1/2)

Dont think thats right

1.2) Hence show that cos(x) = 1-tan^2*1/2(x) / 1+tan^2*1/2(x)

$\displaystyle \cos(2x) = 2\cos^2x - 1$
add 1 to both sides, divide by 2, take square root --> "half-angle" formula

$\displaystyle \cos x = \sqrt{(\cos(2x) + 1)/2}$