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can you help me with trigonometric equations?

cos θ=-sin(-θ)
solve each equation on the interval 0≤θ<2Π
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2 Answers

cos^2 = sin^2
cos^2 = 1-sin^2
1 - sin^2 = sin^2
Add sin^2 to both sides:
1 = 2 sin^2
Divide by 2:
1/2 = sin^2
Take the square root:
+/- sqrt(2)/2 =  sin (-x)
Where is sine +sqrt(2)/2?   pi/4  and 3pi/4
Sine is -sqrt(2)/2 at 5pi/4 and 7pi/4.
cos(pi/4) = -sin(-pi/4)  
cos(5pi/4) = -sin(-5pi/4)
θ = { pi/4, 5pi/4 }


I like Robert's answer better.
Since sin (-x) = -sin (x), you can substitute.
cos x = - (-sin x)
cos x = sin x
Divide by cos x:
cos x = sin x
-----      -----
cos x     cos x
1 = tan x
My way was the longer and much more difficult way.
cosθ = -sin(-θ) is equivalent to
tanθ = 1, since sine is an odd function.
Answer: θ = pi/4, 5pi/4