Let us start subtracting 3 cos theta both sides so that we may get all theta terms on left side only
It is now : 4 cos theta - 3 cos theta = - (sqrt3)/ 2
1 cos theta = -( sqrt 3 ) /2
we know cos (pi-x)= - cos x And cos (pi+ x)= - cos (x)
********* let us first use :
- Cos ( pi - theta) = - ( sqrt 3 )/2
this implies cos (pi- theta )= (sqrt 3) /2
cos ( pi - theta )= cos (pi /6)
pi- theta = pi/6
pi- pi/6 = theta
5 pi/6 = theta
one of teh answer is 5 pi/6 . another value of theta would be :
(2pi - 5pi/6)= 7pi/6
Answer: 5 pi/6 and 7 pi/6
Answer:
Step-by-step explanation:
Rewrite this quadratic equation in standard form: 2n^2 + 3n + 54 = 0. Identify the coefficients of the n terms: they are 2, 3, 54.
Find the discriminant b^2 - 4ac: It is 3^2 - 4(2)(54), or -423. The negative sign tells us that this quadratic has two unequal, complex roots, which are:
-(3) ± i√423 -3 ± i√423
n = ------------------- = ------------------
2(2) 4
Answer:
im sorry but i only know the second one, i believe its 224.3
Step-by-step explanation:
Answer:
the answer is 1
Step-by-step explanation: