Question

Find all solutions in terms of an integer k. Then find the particular solution corresponding to the five value of k? 2sin^2(feta)-1=0 when k=1

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Answer 1

Please refer to the Explanation.

Explanation:

Prerequisite : `sintheta=sinalpha hArr theta=kpi+(-1)^kalpha, k in ZZ`.

Now, `2sin^2theta-1=0 rArr sin^2theta=1/2, or, sintheta=+-1/sqrt2`.

Thus, `sintheta=+-1/sqrt2=sin(+-pi/4)`.

`:. theta=kpi+(-1)^k(+-pi/4), k in ZZ`.

For `k=1, theta=1*pi+(-1)^1(+-pi/4), i.e.,`

`k=1 rArr theta=pi-pi/4=3pi/4, or, theta=pi+pi/4=5pi/4`.

We will enumerate `theta` corresponding to `k=+-1, +-2,& 0`.

`k=1 rArr theta=3pi/4, or, 5pi/4`, as above.

`k=-1:. theta=-pi-pi/4=-5pi/4, or, -pi+pi/4=-3pi/4`.

`k=2:. theta=2pi+-pi/4=9pi/4, or, 7pi/4`.

`k=-2:. theta=-7pi/4, or, -9pi/4`.

`k=0:. theta=+-pi/4`.