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Question: How do you solve \[{\left( {\sin \left( x \right)} \right)^2} = \dfrac{1}{{25}}\] ?...

How do you solve (sin⁡(x))2=125{\left( {\sin \left( x \right)} \right)^2} = \dfrac{1}{{25}} ?

Explanation

Solution

Hint : In order to solve the above trigonometric equation, rewrite the equation by taking the square root on both sides. Derive the solution of sine function first by taking the equation positive in the quadrants 1st and 2nd as sine is always positive in these quadrants and then take the equation as negative and derive two solutions in 3rd and 4th quadrant as sine is negative in these quadrants.

Complete step by step solution:
We are given a trigonometric equation (sin⁡(x))2=125{\left( {\sin \left( x \right)} \right)^2} = \dfrac{1}{{25}} .
(sin⁡(x))2=125{\left( {\sin \left( x \right)} \right)^2} = \dfrac{1}{{25}}
Rewriting the equation by taking square root on both sides. Remember every positive value have two square roots as ±\pm \sqrt {}
sin⁡x=±125\sin x = \pm \sqrt {\dfrac{1}{{25}}}
Putting the value of 25=5\sqrt {25} = 5
sin⁡x=±15\sin x = \pm \dfrac{1}{5} ---(1)
First take the above equation as positive
⇒sin⁡x=15\Rightarrow \sin x = \dfrac{1}{5}
Taking inverse of sine on both sides we have
⇒sin⁡−1(sin⁡x)=sin⁡−1(15)\Rightarrow {\sin ^{ - 1}}\left( {\sin x} \right) = {\sin ^{ - 1}}\left( {\dfrac{1}{5}} \right)
As we can see 15\dfrac{1}{5} is not the remarkable values of sine function , so to find the value of sin⁡−1(15){\sin ^{ - 1}}\left( {\dfrac{1}{5}} \right), take use of the calculator . The value of sin⁡−1(15){\sin ^{ - 1}}\left( {\dfrac{1}{5}} \right)will be sin⁡−1(15)=11.54∘{\sin ^{ - 1}}\left( {\dfrac{1}{5}} \right) = {11.54^ \circ }
x=11.54∘x = {11.54^ \circ }
Since the sine function is positive in 1st and 2nd quadrant both, so using the property of inverse sine function that sin⁡−1(sin⁡x)=π−x{\sin ^{ - 1}}\left( {\sin x} \right) = \pi - x

x=180∘−11∘54 x=168∘46   x = {180^ \circ } - {11^ \circ }54 \\\ x = {168^ \circ }46 \;

Now obtaining the solution by taking the equation (1) negative
⇒sin⁡x=−15\Rightarrow \sin x = - \dfrac{1}{5}
Since the sine function is negative in 3rd and 4th quadrant both, so using the property of inverse sine This will also have two solutions
x=−11∘54x = - {11^ \circ }54
and

x=180∘+11∘54 x=191∘54   x = {180^ \circ } + {11^ \circ }54 \\\ x = {191^ \circ }54 \;

Therefore, the solution to the given trigonometric equation is x=11.54∘x = {11.54^ \circ }, 168∘46{168^ \circ }46 ,−11∘54 - {11^ \circ }54, 191∘54{191^ \circ }54
So, the correct answer is “x=11.54∘x = {11.54^ \circ }, 168∘46{168^ \circ }46 ,−11∘54 - {11^ \circ }54, 191∘54{191^ \circ }54 ”.

Note : 1.One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.
2.The answer obtained should be in a generalised form.
3. Since every positive number has two square roots, one is positive and another is negative. So place the ±\pmfor every positive square root value.
4. The answer obtained is in the interval (0,2π)\left( {0,2\pi } \right).