We know that c o s ( 2 x ) = 1 − 2 s i n 2 ( x ) cos(2x) = 1 - 2 sin^2(x) cos ( 2 x ) = 1 − 2 s i n 2 ( x ) . Now equation looks like
1 − 2 s i n 2 ( x ) + 0.5 s i n ( x ) = 0 1−2sin^2(x) + 0.5 sin(x) = 0 1 − 2 s i n 2 ( x ) + 0.5 s in ( x ) = 0
Multiply the whole equation by -2 (just for convenience):
4 s i n 2 ( x ) − s i n ( x ) − 2 = 0 4 sin^2 (x) - sin(x) -2 =0 4 s i n 2 ( x ) − s in ( x ) − 2 = 0
Now we have a standard quadratic equation for sin(x)
s i n ( x ) = 1 / 8 ( 1 ± 1 + 32 ) = 1 / 8 ( 1 ± 33 ) sin(x) = 1/8 (1 ±\sqrt{1+32}) =1/8 (1 ±\sqrt{33}) s in ( x ) = 1/8 ( 1 ± 1 + 32 ) = 1/8 ( 1 ± 33 )
For the first root (s i n x = 1 / 8 ( 1 + 33 ) sin x = 1/8(1+\sqrt{33}) s in x = 1/8 ( 1 + 33 ) ) the solution isx = ( − 1 ) k a r c s i n ( 1 / 8 ( 1 + 33 ) ) + π k ≈ ( − 1 ) k + π k x = (-1)^k arcsin(1/8(1+\sqrt{33})) + \pi k ≈
(-1)^k + \pi k x = ( − 1 ) k a rcs in ( 1/8 ( 1 + 33 )) + πk ≈ ( − 1 ) k + πk
k =0: x 1 = 1 r a d ≈ 57 ° x_1 = 1 rad ≈ 57° x 1 = 1 r a d ≈ 57°
k = 1: x 2 = − 1 + π = 2.14 r a d ≈ 123 ° x_2 = -1 + \pi = 2.14 rad ≈123° x 2 = − 1 + π = 2.14 r a d ≈ 123°
The same for the second root:
x = ( − 1 ) k a r c s i n ( 1 / 8 ( 1 − 33 ) ) + π k ≈ ( − 1 ) k ∗ ( − 0.635 ) + π k x = (-1)^k arcsin(1/8(1-\sqrt{33})) + \pi k ≈
(-1)^k*(-0.635) + \pi k x = ( − 1 ) k a rcs in ( 1/8 ( 1 − 33 )) + πk ≈ ( − 1 ) k ∗ ( − 0.635 ) + πk
k = 1 : x 3 = ( 0.635 + π ) r a d ≈ 3.78 r a d ≈ 216 ° k =1: x_3= (0.635 + \pi )rad≈3.78 rad ≈ 216° k = 1 : x 3 = ( 0.635 + π ) r a d ≈ 3.78 r a d ≈ 216°
k = 2 : x 4 = ( − 0.635 + 2 π ) r a d ≈ 5.645 r a d ≈ 323 ° k =2: x_4= (-0.635 + 2\pi) rad≈5.645 rad ≈ 323° k = 2 : x 4 = ( − 0.635 + 2 π ) r a d ≈ 5.645 r a d ≈ 323°
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