if sec theta*sin(36° + theta )=0 , find the value of theta so that theta and 36°+theta are acute angles
To find the value of θ such that θ and 36° + θ are acute angles and satisfy the equation sec(θ) * sin(36° + θ) = 0, we can follow these steps:
Step 1: Recognize that sec(θ) * sin(36° + θ) = 0 is a trigonometric equation. To solve it, we need to find the values of θ that make the equation true.
Step 2: Recall that sec(θ) = 1 / cos(θ), so the equation can be rewritten as (1 / cos(θ)) * sin(36° + θ) = 0.
Step 3: Use the Zero Product Property, which states that if a*b = 0, then either a = 0 or b = 0. Apply this property to our equation, setting each factor equal to zero:
a) 1 / cos(θ) = 0
Since the reciprocal of zero is undefined, there is no solution for this part.
b) sin(36° + θ) = 0
Set this factor equal to zero and solve for θ.
Step 4: Consider that sin(θ) = 0 when θ = 0°, 180°, 360°, or any multiple of 180°. Using this property, we can determine the values of θ that satisfy sin(36° + θ) = 0.
sin(36° + θ) = 0
36° + θ = 0°, 180°, 360°, ...
Subtracting 36° from each side, we get:
θ = -36°, 144°, 324°, ...
Step 5: Check if the values of θ we found are acute angles (less than 90°).
-36° is not an acute angle since it is negative.
144° and 324° are both acute angles, as they fall within the range of 0° to 90°.
Step 6: Therefore, the values of θ that satisfy the given conditions (θ and 36° + θ are acute angles) are θ = 144° and θ = 324°.
secØ sin(36+Ø) = 0
secØ = 0 or sin(36°+Ø) = 0
secØ can never be zero, secØ?1 OR secØ ? -1
so:
sin(36°+Ø) = 0
36+Ø = 0° or 180° or 360° or -180° ...
then Ø = -36° ,
Ø = 144° or Ø = 324
since an "acute angle" is defined as an angle between 0° < Ø < 90°
there is no solution to your question.
see Wolfram:
http://www.wolframalpha.com/input/?i=solve+secx+sin(.6283%2Bx)+%3D+0
I changed your variable to x and changed 36° to radians.