From Singapore, Just do not know how to do some question on trigonometry

Question

1) 2cos2θ = -1

2) secθ = 3cosθ

3) secθ = 3cosθ

Thank you so muchhhh!!!!

Answer 1

2cos2T = -1

cos2T =-1/2

reference angle =60

2T = 180 + or - 60

2T = 120 or 240

T = 60 or 120

but 2T also could = 120 + 360 = 480 then T=240

2T could = 240+360 = 600 then T= 300

T = 60, 120, 240 or 300, then add 360n where n= any integer to get an infinite number of solutions

or in radians

T = pi/3, 2pi/3, 4pi/3, or 5pi/3 + 2npi, there's 4 basic solutions. some of these problems limit the answer to 0<T<360, then

T= 60,120,240 or 300

T= 60, 120,240,300,420,480,600, 660,780, ....and the solutions also extend into negative angles

secT = 3cosT

1/cosT =3cosT

cos^2(T) = 1/3

cosT = + or - 1/sqr3 =sqr3/3

about 54.7 degrees is the reference angle

T = about 54.7, or -54.7 degrees + 360n degrees where n= any integer

or about 180 + or - 54.7 = 125.3 or 234.7

if 0<T<360, then

T= about 54.7, 125.3, 234.7 or 315.3 degrees

Answer 2

First of all, problems 2) and 3) are the same.

1) 2cos(2θ)=-1

Let X=2θ. So, 2cos(X)=-1.

Cos(X)=-1/2.

Since cosine of an angle is negative in the 2nd and the 3rd quadrant.

Then X (in radians) ={(2π)/3+2πn, (4π)/3+2πn} where n∈Ζ.

Therefore θ (in radians) ={π/3+2πn, (2π)/3+2πn} where n∈Ζ.

2) secθ=3cosθ

Write the equation in terms of only cosθ(or secθ) and solve. Note secθ=1/cosθ.

So, 1/cosθ=3cosθ becomes 1=3cos^2(θ).

Then solve for cosθ.

cosθ=+-(sqrt(3)/3)

Again, cosine of an angle is positive in the 1st and 4th quadrant and

negative in the 2nd and 3rd quandrant.

Therefore θ(in radians)= {0.96+2πn, (2π-0.96)+2πn, (π-0.96)+2πn, (π+0.96)+2πn} where n∈Ζ.