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8sin²x+sinx−cosx+cos² x=4 solve it?

User Rian
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Final answer:

To solve the equation 8sin²x+sinx-cosx+cos²x=4, we need to isolate sin²x and cos²x separately, and then equate them to find the values of x.

Explaination:

Let's start by isolating sin²x:

8sin²x = 4 - sinx - cosx + cos²x

Dividing both sides by 8:

sin²x = (1/2) - (1/8)sinx - (1/8)cosx + (1/2)cos²x

Now, let's isolate cos²x:

cos²x = (1/2) + (1/8)sinx - (1/8)cosx + sin²x

Next, we substitute the value of sin²x in the second equation:

cos²x = (1/2) + (1/8)sinx - (1/8)cosx + [(1/2) - (1/8)sinx - (1/8)cosx + (1/2)cos²x]

Simplifying:

cos²x = [(3/4) - (3/16)sinx] + [(3/4)cos²x]

Now, we can see that cos²x and cos²x are equal to each other. This means that they must be equal to the same value. Let's call this value k:

cos²x = k = [(3/4) - (3/16)sinx] / [(3/4)]

Squaring both sides:

cos^4 x = k^2 = [(3/4)^2 - (3/16)^2 sin^2 x] / [(3/4)^2]

Finally, we can find the values of x by solving for sin^2 x:

sin^2 x = [(3/4)^2 - k^2] / [(3/16)^2]

Now, let's find the values of k for which this expression is positive and negative. This will give us the range of values for which the equation has real solutions. Here's a table:

| k | Range of x |

| --- | --- |

| 0 | 0° <= x <= 90° or 270° <= x <= 360° |

| (-3/4)/(-3/16) | 90° <= x <= 180° or 360° <= x <= 540° |

| (-3/4)/(-5/16) | 90° <= x <= 270° or 540° <= x <= 720° |

| (-3/4)/(-7/16) | 90° <= x <= 360° or 720° <= x <= 900° |

User Yannickpulver
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