# Thread: 88sin(x) + 242cos(x) = 250

1. ## 88sin(x) + 242cos(x) = 250

solve this using algebra

2. 88sin(x) + 242cos(x) = 250, solve for x

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use this identity: a sin x + b cos x = (a^2 + b^2)^1/2 sin (x + C),

where C = arcsin (b/(a^2 + b^2)), for a is greater or less than 0, also

= pi - arcsin (b/(a^2 + b^2)), for a less than 0.

Then C1 = arcsin (b/(a^2 + b^2))

= arcsin(242/(88^2 + 242^2)^1/2) = arcsin (11/sqrt 137)

= 70.02 degrees

C2 = pi - C1 = 180 - 72.02 degrees = 107.98 degrees

Also, (a^2 + b^2)^1/2 sin (x + C) = (88^2 + 242^2)^1/2) sin (x + C)

= 22 (sqrt 137) sin (x + C)

= 257.5 sin (x + C)

But, 257.5 sin (x + C) = 250, simplifying

sin (x + C) = 250/257.5

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x1 + C1 = arcsin (250/257.5) = 76.14 degrees

x1 = 76.14 - C1

x1 = 76.14 - 70.02 = 6.12 degrees

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or equivalently,

x2 = 76.14 - c2 = 76.14 - 107.98 = -31.84 degrees

3. 88sin(x) + 242cos(x) = 250, solving this using algebra?

what a tall order, a trigonometric function to BE solved by ALGEBRA.

Any HINT?