90% of 10th Graders Can't Solve This!

Added: 2026-05-26

[00:00:00]These are the types of questions I'm paying my internet bills for as it is actually interesting and it will consume a bit of our brain cells. We have 81 raised to sin square x plus 81 raised to cos square x= 30. Our job is to find the values of x which lies in this range 0 to 2 pi radians or 0 to 360°.

[00:00:27]So can you solve it? Okay, at first glance we might think that this question cannot be solved using algebra. But if you know a bit about power rules, quadratic equations and this one thing in trigonometry, then this question will become a piece of cake. Before we begin, let us first rewrite this 81 as 3 to the^ 4. So the equation will become this plus this equals 30. We have just substituted 81 as 3 to 4th. Now using this power rule, we can rewrite this as 3 raised to the 4th power * sin^ 2 x plus this as 3 raised to the 4th power * cos^² x = 30.

[00:01:15]Now I was talking about this trigonometric identity. We all know this, right? We can express cos^² x as 1 - sin^ 2 x. This gives us 3 ra 4 * sin^ 2 x + 3 raised to the 4 * 1 - sin^ 2 x = 30 or this can also be written as 3 raised to 4 - 4 * sin^ 2 x. Let's call this 4 * sin^ 2 x as r.

[00:01:48]So we now have 3 raised to the r + 3 ra to 4 - r = 30. Now let's multiply the entire equation by 3 ra to r and see the magic. This will become 3 r^ 2. Now using this power rule we get this as 3 4 - r + r or 3 4 or 81 and this equals 30 * 3 to the r noise. Now let us substitute p as 3 to the r. So this equation will become p² + 81 = 30 p.

[00:02:29]Awesome. At least now we know how to solve this quadratic equation. I will not bore you by solving this equation.

[00:02:38]So bang p = 27 and 3 and we have p = 3 to the r.

[00:02:46]Now when p = 3, 3 to the r = 3 gives r = 1 and when p = 27, 3 to the r = 27, which is also equal to 3 cub. Comparing the powers and bases we get r= 3. Right?

[00:03:06]So we have r = 1 and 3. Now we go back to our original trigonometric equation.

[00:03:13]We have r = 4 * sin^ 2 x. So if r is 3, 4 * sin^ 2 x = 3, which gives us sin^ 2 x = 3 / 4.

[00:03:28]And if r is 1 then 4 * sin^ 2 x = 1 which gives us sin^ 2 x = 1 divided by 4. Finally to find the values of x we take the square root of both sides. Sin x = plus or minus the square<unk> of 3 / 2. And for sin^ square x = 1 / 4 we get sin x = + or minus 12. So we are looking for the possible values of x where the s of x equals either plus or minus the<unk> of 3 / 2 or plus or minus 1/2 within the range from 0 to 2<unk>i radians or 0 to 360°.

[00:04:18]To solve for x, we will look at the graph of sine of x from 0 to 360°.

[00:04:25]First, the sine of x equals the square<unk> of 3 / 2 at two values. In degrees, these values are x= 60 and 120.

[00:04:38]Next, the s of x= minus the<unk> of 3 / 2 at two values in degrees. These values are x= 240 and 300. Now for the s of x = 12. In degrees these values are x= 30 and 150.

[00:05:00]Lastly the s of x = - 1/2 at two values in degrees. These values are x= 210 and 330. So all the possible values for x are boom right here. Isn't this super cool? If you enjoyed this video, please don't forget to like, share, and subscribe to our channel. So, good.

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