Selesaikan 2cos2θ+1=0 | Microsoft Math Solver

Cari nilai θ

θ = π n_{1} + \frac{2 π}{3}, n_{1} \in Z

θ = π n_{1} + \frac{π}{3}, n_{1} \in Z

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5 soal serupa dengan:

2 cos 2 θ + 1 = 0

Soal yang Mirip dari Pencarian Web

How do you solve

2 cos θ + 1 = 0

https://socratic.org/questions/how-do-you-solve-2costheta-1-0

The general solution of

2 cos θ + 1 = 0

θ = 2 k π \pm \frac{2 π}{3}, k \in Z

Explanation: Here,

2 cos θ + 1 = 0

How do you use product to sum formulas to write the product

cos 5 θ cos 3 θ

as a sum or difference?

https://socratic.org/questions/how-do-you-use-product-to-sum-formulas-to-write-the-product-cos5thetacos3theta-a

cos 5 θ cos 3 θ = \frac{1}{2} cos 8 θ + \frac{1}{2} cos 2 θ

Explanation: As

cos (A + B) = cos A cos B - sin A sin B

How do you find the solution to

5 (cos θ + 1) = 5

0 \leq θ < 360

https://socratic.org/questions/how-do-you-find-the-solution-to-5-costheta-1-5-if-0-theta-360

\frac{π}{2}

\frac{3 π}{2}

Explanation: I assume you know

0 \leq θ < 360

is the same thing as

0 \leq θ < 2 π

csc^{2} (\frac{π}{9}) + csc^{2} (\frac{2 π}{9}) + csc^{2} (\frac{4 π}{9})

https://math.stackexchange.com/q/1749306

Like my answer here in Sum of tangent functions where arguments are in specific arithmetic series

tan 9 x = \frac{( 1 9 ) t - ( 3 9 ) t ^{3} + ( 5 9 ) t ^{5} - ( 7 9 ) t ^{7} + t ^{9}}{\dots}

t = tan x

tan 9 x = 0, 9 x = n π

Exact period of simple pendulum.

https://math.stackexchange.com/questions/2257095/exact-period-of-simple-pendulum

- g sin θ = L θ^{''} = L \frac{d ^{2} θ}{d t ^{2}}

- g sin θ \frac{d θ}{d t} = L \frac{d ^{2} θ}{d t ^{2}} \frac{d θ}{d t}

- 2 g sin θ d θ = L \times 2 \frac{d ^{2} θ}{d t ^{2}} \frac{d θ}{d t} d t = L [(\frac{d θ}{d t})^{2}]^{'} d t

Polar equation — find area under graph using double integral

https://math.stackexchange.com/questions/729893/polar-equation-find-area-under-graph-using-double-integral

You have the heuristic (and graphic) result: http://www.mathalino.com/sites/default/files/images/plane-areas-polar-coordinates.jpg the part of the graphic in color is close to a triangle with both ...

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