sin 2 theta + cos theta = 0
so, we use sin 2 theta = 2 sin theta cos theta right? but the problem is i'm stuck on (2 sin theta cos theta) + cos theta = 0. What should i do?
cosTheta(2SinTheta+1)=0
remember ab=o? a or b or both is = zero
i see you factorised the question. But, still i can't find the solution. but how about this:
2sin theta cos theta = - cos theta
2sin theta = - cos theta/cos theta
2sin theta = -1
sin theta = -1/2
theta = sin^-1(1/2)
theta = 30
from Bob's , cosØ(2sinØ + 1)=0
cosØ = 0 or sinØ = -1/2
if cosØ = 0, then Ø = 90° or 270° OR Ø = π/2 , 3π/2
if sinØ = -1/2, Ø must be in III or IV
Ø = 180+30 = 210° or 360-30 = 330°
in radians Ø = 7π/6 , 11π/6
you totally ignored the fact it was -1/2
and did not take the CAST rule into consideration.
Yeah, usually i used sinØ = -1/2 to find the quarters and skipped the negative to find theta.
"if sinØ = -1/2, Ø must be in III or IV. Hence, Ø = 180+30 = 210° or 360-30 = 330°"
However, i got the same answers as yours :). Just to make sure whether it is correct or not.
Thanks for the guide bobpursley and Reiny!
To solve the equation sin 2 theta + cos theta = 0, you correctly identified that you can use the identity sin 2 theta = 2 sin theta cos theta. Now you are left with the equation (2 sin theta cos theta) + cos theta = 0.
To simplify this equation, you can factor out the common factor of cos theta:
cos theta * (2 sin theta + 1) = 0.
Now you have two possibilities for the equation to be true:
1. cos theta = 0: This means that theta is equal to (2n + 1) * pi/2, where n is an integer.
2. 2 sin theta + 1 = 0:
Subtracting 1 from both sides gives you:
2 sin theta = -1.
Dividing by 2 gives you:
sin theta = -1/2.
To find the values of theta that satisfy sin theta = -1/2, you can use the unit circle or trigonometric ratios of special angles. The angles in the unit circle where sin theta = -1/2 are theta = 7pi/6, 11pi/6, and their coterminal angles.
So, the solutions for theta are:
theta = (2n + 1) * pi/2, 7pi/6 + 2pi * k, 11pi/6 + 2pi * k,
where n and k are integers.