Math  /  Trigonometry

QuestionSolve cos2(x)=8sin(x)\cos ^{2}(x)=-8 \sin (x) for all solutions 0x<2π0 \leq x<2 \pi. x=x=

Studdy Solution
Evaluate the solutions for sin(x)\sin(x):
1. sin(x)=4+17\sin(x) = 4 + \sqrt{17} is not possible since the sine function's range is [1,1][-1, 1].
2. sin(x)=417\sin(x) = 4 - \sqrt{17}

Calculate 4174 - \sqrt{17}:
41744.123=0.1234 - \sqrt{17} \approx 4 - 4.123 = -0.123
Since 0.123-0.123 is within the range [1,1][-1, 1], it is a valid solution for sin(x)\sin(x).
Find xx such that sin(x)=0.123\sin(x) = -0.123 in the interval 0x<2π0 \leq x < 2\pi:
1. First solution: x=arcsin(0.123)x = \arcsin(-0.123)
2. Second solution: x=πarcsin(0.123)x = \pi - \arcsin(-0.123)

Calculate:
x1=arcsin(0.123)0.123x_1 = \arcsin(-0.123) \approx -0.123
Adjust to the interval 0x<2π0 \leq x < 2\pi:
x1=2π0.1236.160x_1 = 2\pi - 0.123 \approx 6.160
Calculate the second solution:
x2=π(0.123)=π+0.1233.265x_2 = \pi - (-0.123) = \pi + 0.123 \approx 3.265
The solutions for xx in the interval 0x<2π0 \leq x < 2\pi are:
x3.265,6.160 x \approx 3.265, 6.160

View Full Solution - Free
Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord