Answer:x=(n−12)π or (2n−1π)
Explanation:
-4 sin 2 x−10(sin x−1)(sin+)=sin x−=0 sin x=1 x 1π5π...x,x−,=2,...
we can say that,
x=12π,32π,52π,72π,...
we can use arithmetic progression to solve this sequence where,
a=12π,d=π.Tn=12π+(n−1)π=(12+n−1)πTn=(n−12)π or (2n−1)π2
Therefore it general solution for
x is (n 12π or (n−1)π2 for all n
Hope this helps!!!
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