Area A=4sinθ + 2sin2θ,where 0<θ< pi/2.Find the two possible values of θ, when an area A is 5m^2? On the other hand, if area A is the same but different values of  θ, find all possible...

Area A=4sinθ + 2sin2θ,where 0<θ< pi/2.

Find the two possible values of θ, when an area A is 5m^2? On the other hand, if area A is the same but different values of  θ, find all possible values for A ?

Asked on by nicoleuc

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to substitute 5 for A in equation `A = 4 sin theta + 2 sin 2theta`  such that:

`5 =4sin theta + 2 sin 2theta`

You need to solve for `theta`  the equation above, thus you need to express the equation in terms of `theta`  using the formula of double angle such that:

`5 = 4sin theta + 2 (2 sin theta*cos theta)`

`5 = 4sin theta + 4sin theta*cos theta`

You need to factor out `theta ` such that:

`5 = 4sin theta(1 + cos theta)`

You need to divide by 4 both sides such that:

`5/4 = sin theta(1 + cos theta)`

`1.25 = sin theta(1 + cos theta)`

You should write `1.25 = 1*1.25`  such that:

`1*1.25 = sin theta(1 + cos theta)`

`sin theta = 1 =gt theta = pi/2`

Notice that `theta = pi/2`  is not a convenient value since `theta`  needs to be smaller than `pi/2` .

`cos theta + 1 = 1.25 =gt cos theta = 1.25 - 1`

`cos theta = 0.25 =gt theta = cos^(-1) 0.25`

`theta ~~ 75.52^o`

Notice that you may only select `sin theta = 1` , since `sin theta = 1.25 ` becomes a contradiction because the values of sine function are not larger than 1.

Hence, evaluating the solutions to equation yields that `theta`  has only one convenient value of `75.52^o` .

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