Exact stationary points of f(x) = e^(2x) • sin^(2x) 0</=x</= 1/2pi

[raw socket]

I’m able to use the product rule uv’+u’v to find the derivative of the function which is e^(2x) • 2cos2x + 2e^(2x) • sin^(2x) then set it equal to zero to find the coordinates of the stationary points ~ I’ve spent a lot of time trying to solve for x but still can’t get it! I’m guessing the solution involves the double angle formulas or substitution but I’m stuck pls help 🥹

[random cove]

Is the function $e^{2x}\cdot{sin{2x}}$

[kind orchid]

Victimizer

[random cove]

Also u uan solve it without double angle formula

Take $2e^{2x}$ common from both terms

[kind orchid]

Victimizer

[random cove]

And then u get $sin{2x}+cos{2x}$[only if what i considered was the function, right]

[kind orchid]

Victimizer

Victimizer

[random cove]

And u can get a value of x from there

U can also say
$sin{2x} +cos{2x}=0$

[kind orchid]

Victimizer

[random cove]

Subtracting cos2x from both sides ,u can square both side and convert one either of them into the other one using sin²x +cos²x=1
and then u can solve for x

Then whatver values u get for x ,check if they r within the given interval,if they r
Yay we got it
Otherwise choose the one which is within the interval

I think u will get 2 values and 1 of them will be within the interval

If u dont understand, u can ask

[charred linden]

bruh

This is giberish

[raw socket]

@random cove Thanks for your explanation! I didn’t even think about 2e^2x cannot be equal = 0 😅😅

[random cove]

Np

!done

[thorn badger]

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[raw socket]

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