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How do you find the exact value of #4sin^2theta-3sintheta=1# in the interval #0<=theta<2pi#?

Answer 1

Zoe Austin

#theta={pi/2,3.39,6.03}#

#4sin^2theta-3sintheta=1#

#hArr4sin^2theta-3sintheta-1=0#

or #4sin^2theta-4sintheta+sintheta-1=0#

or #4sintheta(sintheta-1)+1(sintheta-1)=0#

or #(4sintheta+1)(sintheta-1)=0#

therefore either #4sintheta+1=0# i.e. #sintheta=-1/4# and #theta=345.52^@=6.03# or #194.48^@=3.39#

or #sintheta=1# i.e. #theta=pi/2#

Hence #theta={pi/2,3.39,6.03}#

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Answer 2

Zoe Austin

To find the exact value of (4\sin^2(\theta) - 3\sin(\theta) = 1) in the interval (0 \leq \theta < 2\pi), you can rewrite it as a quadratic equation in terms of (\sin(\theta)). Let (x = \sin(\theta)), then solve the resulting quadratic equation (4x^2 - 3x - 1 = 0). Once you find the values of (x), substitute them back into (\sin(\theta)) to find the corresponding values of (\theta) within the given interval.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you find the exact value of #4sin^2theta-3sintheta=1# in the interval #0&lt;=theta&lt;2pi#?

How do you find the exact value of #4sin^2theta-3sintheta=1# in the interval #0<=theta<2pi#?