solve equations on the interval

Question

Solve each equation on the interval 0<Θ<2pi

 

a) 1-sin(Θ)=1/2

b) 2sin^2(Θ)-1=0

c) tan2(Θ) =-1

d) cos3(Θ)=1

Answer 1

tan(2ø)= -1  let u=2ø so we get tan u = -1. This has an answer in quadrant 2and 4. tan 45 =1 so we have a 45 degree reference angle. so that leads to 135 and 315. The TRICKY part of double or triple angle problems is that you need coterminal angles, two for double angles,three for triple angle .... so

135 and 495 (135+360) and 315 and 675 (315+360) so we have 

2ø=135      or ø=67.5

2ø=495      or ø=247.5

2ø=315      or ø=157.5

2ø=675      or ø=337.5

 

 

 

cos 3ø = 1 let u=3ø so cos u = 1 which happens u=0. you need three coterminal angles so 0, 360, 720.

so 3ø = 0, ø=0

3ø = 360, so ø=120

3ø=720, so ø=240

Answer 2

 

To get you started...

 

a) 1-sin(x)=1/2

    -sin(x)=1/2-1

    -sin(x)=-1/2

     sin(x)=1/2

     sin is positive in the 1st and 2nd quadrants

     sin(30º)=1/2

     sin(150º)=1/2 (make a 30º angle with the negative x-axis and now 180-30=150)

 

b) 2sin²(x)-1=0

    2sin²(x)=1

    sin²(x)=1/2

    sin²(x)=[1-cos(2x)]/2

    substitute

    1/2=[1-cos(2x)]/2

    the denominators are equal so set the numerators equal

    1=1-cos(2x)

    0=-cos(2x)

    cos(2x)=0

    cos(90º)=0 implies 2x=90 and x=45º

    cos(270º)=0 implies 2x=270 and x=135º

 

 c) tan2(x)=-1

     tan is negative in 2nd and 4th quadrants

     tan2(x)=2tan(x)/(1-tan²(x))

     2tan(x)/(1-tan²(x))=-1

     2tan(x)=tan²(x)-1

     tan²(x)-2tan(x)-1=0

     let y=tan(x)

     y²-2y-1=0

     using the quadratic equation...

     (2+√8)/2

     (2+2√2)/2

     1+√2=y

     1+√2=2.41421356

     1-√2=-0.41421356

     tan(?)=2.414...

     ?=67.5º, so x=135º

     tan(?)=-0.414...

     ?=157.5º, so x=315º

     another solution...

     tan=sin/cos

     when are sin and cos additive inverses ?

     looking on the unit circle...

     at 135º(sin=√2/2 and cos=-√2/2) and...

     at 315º(sin=-√2/2 and cos=√2/2)