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Question: How do you solve \[{{\sin }^{2}}x=0\] and find all solutions in the interval \[0\le x<360\]?...

How do you solve sin⁡2x=0{{\sin }^{2}}x=0 and find all solutions in the interval 0≤x<3600\le x<360?

Explanation

Solution

In this problem, we have to solve and find the value of x in the given trigonometric expression within the given interval. We can write the given trigonometric expression as the whole square form and we can take the square on both sides to get a simplified form of the given expression. We can write the values of x for which the sin value becomes 0, hence we can solve for x from the given trigonometric expression. We can write the answer either in degree or in radians.

Complete step by step answer:
We know that the given trigonometric expression to be solved is,
sin⁡2x=0{{\sin }^{2}}x=0 within the interval 0≤x<3600\le x<360.
We can write the above trigonometric expression as,
⇒sin⁡2x=(sin⁡x)2\Rightarrow {{\sin }^{2}}x={{\left( \sin x \right)}^{2}}
Now we can write the expression,
⇒(sin⁡x)2=0\Rightarrow {{\left( \sin x \right)}^{2}}=0
Now we can take square on both sides we get
⇒sin⁡x=0\Rightarrow \sin x=0
We know that when x is 0 and 180∘{{180}^{\circ }} within the given interval 0≤x<3600\le x<360, then the sin value becomes 0.
Therefore, the value of x = 0, 180∘{{180}^{\circ }}.

Note:
Students make mistakes while simplifying the given expression to solve for x, which should be concentrated. We should know trigonometric identities, formulas and degree values to solve these types of problems. In this problem, we have used sin values which is 0, which we should know to solve this problem. We can also write the answer in radian form, instead of degree form.